Step 1: Identifying the Indeterminate Form

Given the expression: $L = \lim_{x \to 0} \frac{\tan(x) - x}{x^2 \tan(x)}$.

Substituting $x = 0$ into the expression: $\tan(0) = 0$. Thus, $L = \frac{0 - 0}{0^2 \cdot 0} = \frac{0}{0}$. [This is an indeterminate form, allowing the application of L'Hôpital's Rule or Taylor Series expansion].

Step 2: Taylor Series Expansion Method

To evaluate the limit efficiently, we expand $\tan(x)$ using its Taylor series about $x=0$:

$\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$

Step 3: Substituting the Expansion into the Numerator and Denominator

Numerator: $\tan(x) - x = (x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots) - x = \frac{x^3}{3} + \frac{2x^5}{15} + \dots$

Denominator: $x^2 \tan(x) = x^2(x + \frac{x^3}{3} + \dots) = x^3 + \frac{x^5}{3} + \dots$

Step 4: Algebraic Simplification

Now, rewrite the limit as $x \to 0$:

$L = \lim_{x \to 0} \frac{\frac{x^3}{3} + \frac{2x^5}{15} + \dots}{x^3 + \frac{x^5}{3} + \dots}$

Divide both the numerator and the denominator by $x^3$ [By the property of limits for algebraic functions]:

$L = \lim_{x \to 0} \frac{\frac{1}{3} + \frac{2x^2}{15} + \dots}{1 + \frac{x^2}{3} + \dots}$

Step 5: Evaluating the Limit

As $x \to 0$, all terms containing $x$ vanish to zero:

$L = \frac{\frac{1}{3} + 0}{1 + 0} = \frac{1}{3}$

The graph visualizes the behavior of the numerator near the origin, showing it approaches $0$ at a rate proportional to $x^3$.

Final Solution: \frac{1}{3}

Step 1: Formal Definition and Substitution

To evaluate the limit $L = \lim_{x \to 0} \frac{(1+x)^{1/6} - 1}{(1+x)^{1/21} - 1}$, we observe that direct substitution yields an indeterminate form of type $\frac{0}{0}$. We employ the substitution $u = 1+x$. As $x \to 0$, it follows that $u \to 1$.

Substituting $u$ into the expression:

$L = \lim_{u \to 1} \frac{u^{1/6} - 1}{u^{1/21} - 1}$

Step 2: Applying the Fundamental Limit Theorem

We utilize the standard algebraic limit formula: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$ [By the Binomial Limit Theorem].

To manipulate our expression into this form, we divide both the numerator and the denominator by $(u - 1)$:

$L = \lim_{u \to 1} \frac{\frac{u^{1/6} - 1}{u - 1}}{\frac{u^{1/21} - 1}{u - 1}}$

Step 3: Calculating Limits of Numerator and Denominator

Applying the quotient rule for limits [Per Limit Laws]:

Numerator: $\lim_{u \to 1} \frac{u^{1/6} - 1^{1/6}}{u - 1} = \frac{1}{6}(1)^{(1/6) - 1} = \frac{1}{6}$

Denominator: $\lim_{u \to 1} \frac{u^{1/21} - 1^{1/21}}{u - 1} = \frac{1}{21}(1)^{(1/21) - 1} = \frac{1}{21}$

Step 4: Final Evaluation

Substituting the results from Step 3:

$L = \frac{1/6}{1/21} = \frac{21}{6}$

Simplifying the fraction by dividing both terms by the greatest common divisor, $3$:

$L = \frac{7}{2}$

Final Solution: \frac{7}{2} \text{ or } 3.5

Given Function

Let the function be defined as $f(x) = (3x + 5)(1 + \tan(x))$.

Step 1: Application of the Product Rule

To differentiate the product of two functions, we employ the Product Rule of differentiation. For $f(x) = u(x) \cdot v(x)$, the derivative is defined as [Per the Product Rule]:

$$\frac{d}{dx}[u(x)v(x)] = u(x) \frac{d}{dx}[v(x)] + v(x) \frac{d}{dx}[u(x)]$$

Let $u(x) = (3x + 5)$ and $v(x) = (1 + \tan(x))$.

Step 2: Calculating Individual Derivatives

We derive the components separately:

• For $u(x) = 3x + 5$: $$\frac{d}{dx}(3x + 5) = \frac{d}{dx}(3x) + \frac{d}{dx}(5) = 3 + 0 = 3$$ [By the Sum Rule and Power Rule]
• For $v(x) = 1 + \tan(x)$: $$\frac{d}{dx}(1 + \tan(x)) = \frac{d}{dx}(1) + \frac{d}{dx}(\tan(x)) = 0 + \sec^2(x) = \sec^2(x)$$ [By the derivative of trigonometric functions]

Step 3: Substitution and Expansion

Substituting these results back into the Product Rule formula:

$$f'(x) = (3x + 5) \cdot \sec^2(x) + (1 + \tan(x)) \cdot 3$$

Distribute the terms to simplify the expression:

$$f'(x) = (3x + 5)\sec^2(x) + 3 + 3\tan(x)$$

Step 4: Final Algebraic Arrangement

The derivative is consolidated as:

$$f'(x) = (3x + 5)\sec^2(x) + 3\tan(x) + 3$$

Final Solution: \frac{d}{dx}[(3x + 5)(1 + \tan(x))] = (3x + 5)\sec^2(x) + 3\tan(x) + 3

Step 1: Identifying the Mathematical Form

The expression to evaluate is defined as:

$L = \lim_{x \to 0} \frac{(1+x)^n - 1}{x}$

Direct substitution of $x = 0$ yields:

$\frac{(1+0)^n - 1}{0} = \frac{1^n - 1}{0} = \frac{0}{0}$

This indicates an indeterminate form, necessitating the application of the Binomial Theorem or L'Hôpital's Rule [By Indeterminate Form Theory].

Step 2: Applying the Binomial Theorem

For any rational $n$, the expansion of $(1+x)^n$ where $|x| < 1$ is given by the Binomial series [By Binomial Theorem]:

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

Step 3: Algebraic Substitution and Simplification

Substitute the expansion into the limit expression:

$L = \lim_{x \to 0} \frac{\left( 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \right) - 1}{x}$

Subtracting $1$ from the numerator:

$L = \lim_{x \to 0} \frac{nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots}{x}$

Step 4: Division by $x$

Factoring $x$ from the numerator:

$L = \lim_{x \to 0} \left[ \frac{x \left( n + \frac{n(n-1)}{2!}x + \frac{n(n-1)(n-2)}{3!}x^2 + \dots \right)}{x} \right]$

Canceling the term $x$ (since $x \to 0$ but $x \neq 0$):

$L = \lim_{x \to 0} \left( n + \frac{n(n-1)}{2!}x + \frac{n(n-1)(n-2)}{3!}x^2 + \dots \right)$

Step 5: Evaluating the Limit

As $x \to 0$, all terms containing $x$ vanish [By Limit Properties]:

$L = n + 0 + 0 + \dots = n$

Final Solution: \lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n

Step 1: Analyzing the Expression and Identification of Indeterminate Forms

Given the expression: $f(x) = \frac{8x-3}{2x-1} - \frac{4x^2+1}{4x^2-1}$

We are required to evaluate $\lim_{x \to \frac{1}{2}} f(x)$.

First, observe the denominator of the second term: $4x^2 - 1$. By the [Difference of Two Squares Identity], $a^2 - b^2 = (a-b)(a+b)$, we can factorize it as: $4x^2 - 1 = (2x - 1)(2x + 1)$

Step 2: Finding a Common Denominator

Rewrite the expression to facilitate subtraction:

$f(x) = \frac{8x-3}{2x-1} - \frac{4x^2+1}{(2x-1)(2x+1)}$

To subtract, find the Lowest Common Denominator (LCD), which is $(2x-1)(2x+1)$. Multiply the numerator and denominator of the first term by $(2x+1)$:

$f(x) = \frac{(8x-3)(2x+1) - (4x^2+1)}{(2x-1)(2x+1)}$

Step 3: Algebraic Expansion and Simplification

Expand the numerator:

$(8x-3)(2x+1) = 16x^2 + 8x - 6x - 3 = 16x^2 + 2x - 3$

Substitute this back into the expression:

$f(x) = \frac{(16x^2 + 2x - 3) - (4x^2 + 1)}{(2x-1)(2x+1)}$ $f(x) = \frac{16x^2 - 4x^2 + 2x - 3 - 1}{(2x-1)(2x+1)}$ $f(x) = \frac{12x^2 + 2x - 4}{(2x-1)(2x+1)}$

Step 4: Factorizing the Numerator

Factorize $12x^2 + 2x - 4$. Factor out the common constant $2$:

$2(6x^2 + x - 2)$

Using the splitting-the-middle-term method for $6x^2 + x - 2$: find two numbers that multiply to $(6 \times -2) = -12$ and add to $1$. These numbers are $4$ and $-3$.

$6x^2 + 4x - 3x - 2 = 2x(3x + 2) - 1(3x + 2) = (2x - 1)(3x + 2)$

Thus, the expression becomes:

$f(x) = \frac{2(2x - 1)(3x + 2)}{(2x - 1)(2x + 1)}$

Step 5: Evaluating the Limit

For $x \neq \frac{1}{2}$, the term $(2x-1)$ is non-zero and can be cancelled [By the limit definition, as $x$ approaches $a$, $x \neq a$]:

$\lim_{x \to \frac{1}{2}} f(x) = \lim_{x \to \frac{1}{2}} \frac{2(3x + 2)}{2x + 1}$

Since the denominator $(2x+1)$ is non-zero at $x = \frac{1}{2}$, substitute the value directly:

$\text{Limit} = \frac{2(3(\frac{1}{2}) + 2)}{2(\frac{1}{2}) + 1}$ $\text{Limit} = \frac{2(\frac{3}{2} + 2)}{1 + 1} = \frac{2(\frac{7}{2})}{2}$ $\text{Limit} = \frac{7}{2}$

Visual Representation of Function Behavior

Final Solution: \frac{7}{2}

Step 1: Evaluation of $\lim_{x \to -2} f(x)$

The definition of the function $f(x)$ is provided for the interval $0 < x < 3$. We are asked to evaluate $\lim_{x \to -2} f(x)$. However, based on the domain of the piecewise function $f(x)$, which is defined only for $x \in (0, 3)$, the value $x = -2$ lies outside the domain of the function.

[Per the definition of limits in real analysis, the limit of a function at a point $a$ exists only if the function is defined in a punctured neighborhood of $a$. Since $f(x)$ is not defined in any neighborhood of $x = -2$, the limit $\lim_{x \to -2} f(x)$ does not exist.]

Correction/Assumption: In standard academic contexts for this specific problem type, if a limit evaluates to a point outside the domain, we investigate if there is a typographical error in the problem statement (e.g., $x \to 0^+$). If we evaluate $\lim_{x \to 0^+} f(x)$ instead:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 1) = (0)^2 - 1 = -1$.

Step 2: Evaluation of $\lim_{x \to 2^+} f(x)$

To find the right-hand limit at $x = 2$, we must use the sub-function defined for the interval $x \geq 2$.

For $x \geq 2$, $f(x) = 2x + 3$.

$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x + 3)$

Substituting $x = 2$ directly into the expression [By the Direct Substitution Property for polynomial functions]:

$= 2(2) + 3 = 4 + 3 = 7$.

Step 3: Constructing the Quadratic Equation

Given the roots of the quadratic equation are $\alpha = -1$ and $\beta = 7$, we use the standard form of a quadratic equation given its roots:

$x^2 - (\alpha + \beta)x + (\alpha \beta) = 0$

Substitute the values of the roots:

Sum of roots $(\alpha + \beta) = -1 + 7 = 6$

Product of roots $(\alpha \beta) = (-1)(7) = -7$

Step 4: Algebraic Formulation

Substituting the sum and product into the quadratic template:

$x^2 - (6)x + (-7) = 0$

$x^2 - 6x - 7 = 0$

Final Solution: The quadratic equation is $x^2 - 6x - 7 = 0$

Given Function

Let the function be defined as $f(x) = \frac{1}{ax^2 + bx + c}$.

Step 1: Reformulation for Differentiation

Rewrite the function using negative exponent notation to facilitate the application of the Generalized Power Rule (Chain Rule):

$f(x) = (ax^2 + bx + c)^{-1}$

[Justification: Per the laws of exponents, $\frac{1}{u^n} = u^{-n}$]

Step 2: Applying the Chain Rule

To find the derivative $\frac{d}{dx}[f(x)]$, let $u = ax^2 + bx + c$. Then $f(u) = u^{-1}$.

The Chain Rule states that $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$.

[Justification: Chain Rule for differentiation of composite functions]

Step 3: Calculating Partial Derivatives

1. Differentiating the outer function $u^{-1}$ with respect to $u$:

$\frac{d}{du}(u^{-1}) = -1 \cdot u^{-1-1} = -u^{-2} = -\frac{1}{u^2}$

[Justification: Power Rule $\frac{d}{dx}(x^n) = nx^{n-1}$]

2. Differentiating the inner function $u = ax^2 + bx + c$ with respect to $x$:

$\frac{d}{dx}(ax^2 + bx + c) = \frac{d}{dx}(ax^2) + \frac{d}{dx}(bx) + \frac{d}{dx}(c)$

$= a(2x) + b(1) + 0 = 2ax + b$

[Justification: Sum Rule and linearity of the derivative operator]

Step 4: Combining and Simplifying

Substitute the partial derivatives back into the chain rule identity:

$\frac{d}{dx}f(x) = -\frac{1}{(ax^2 + bx + c)^2} \cdot (2ax + b)$

$\frac{d}{dx}f(x) = -\frac{2ax + b}{(ax^2 + bx + c)^2}$

Visual Representation of the Slope Function

Final Solution: $-\frac{2ax + b}{(ax^2 + bx + c)^2}$

Step 1: Algebraic Expansion of the Function

Given the function $f(x) = (x + \frac{1}{x})^3$. To differentiate this expression, we may either apply the Chain Rule directly or expand the expression using the binomial theorem, $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Expanding $f(x)$:

$f(x) = x^3 + 3(x^2)(\frac{1}{x}) + 3(x)(\frac{1}{x^2}) + (\frac{1}{x})^3$

Simplifying the terms:

$f(x) = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}$

Expressing as powers of $x$ for differentiation [Per the Power Rule for exponents]:

$f(x) = x^3 + 3x + 3x^{-1} + x^{-3}$

Step 2: Applying the Sum and Difference Rule

The derivative of a sum of functions is the sum of their individual derivatives [Per the Linearity of the Differentiation Operator]:

$\frac{d}{dx}[f(x)] = \frac{d}{dx}(x^3) + \frac{d}{dx}(3x) + \frac{d}{dx}(3x^{-1}) + \frac{d}{dx}(x^{-3})$

Step 3: Calculating Derivatives via the Power Rule

Applying the Power Rule, $\frac{d}{dx}(x^n) = nx^{n-1}$:

• $\frac{d}{dx}(x^3) = 3x^{3-1} = 3x^2$
• $\frac{d}{dx}(3x) = 3(1)x^{1-1} = 3$
• $\frac{d}{dx}(3x^{-1}) = 3(-1)x^{-1-1} = -3x^{-2} = -\frac{3}{x^2}$
• $\frac{d}{dx}(x^{-3}) = -3x^{-3-1} = -3x^{-4} = -\frac{3}{x^4}$

Step 4: Consolidating the Derivative

Combining the results from Step 3:

$f'(x) = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Visual Representation of Function Slope

Step 5: Alternative Verification via Chain Rule

Let $u = x + \frac{1}{x}$. Then $f(u) = u^3$.

$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = 3u^2 \cdot \frac{d}{dx}(x + x^{-1})$ $\frac{df}{dx} = 3(x + \frac{1}{x})^2 \cdot (1 - x^{-2})$ $\frac{df}{dx} = 3(x^2 + 2 + \frac{1}{x^2}) \cdot (1 - \frac{1}{x^2})$ $\frac{df}{dx} = 3(x^2 + 2 + \frac{1}{x^2} - 1 - \frac{2}{x^2} - \frac{1}{x^4})$ $\frac{df}{dx} = 3(x^2 + 1 - \frac{1}{x^2} - \frac{1}{x^4})$ $\frac{df}{dx} = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Final Solution: $f'(x) = 3x^2 + 3 - \frac{3}{x^2} - \frac{3}{x^4}$

Step 1: Identifying the Mathematical Form

To evaluate the limit $L = \lim_{x \to -3} \frac{x^3 + 27}{x^5 + 243}$, we first perform direct substitution to check for indeterminacy.

Substituting $x = -3$:

Numerator: $(-3)^3 + 27 = -27 + 27 = 0$

Denominator: $(-3)^5 + 243 = -243 + 243 = 0$

Since the form is $\frac{0}{0}$, the expression is indeterminate. We must resolve this by applying the standard limit theorem: $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ [Per the Standard Limit Theorem for Algebraic Functions].

Step 2: Algebraic Manipulation of the Expression

To utilize the theorem, we rewrite both the numerator and denominator in the form $x^n - a^n$. Here, $a = -3$.

Numerator: $x^3 + 27 = x^3 - (-27) = x^3 - (-3)^3$

Denominator: $x^5 + 243 = x^5 - (-243) = x^5 - (-3)^5$

Now, rewrite the limit as follows:

$L = \lim_{x \to -3} \frac{x^3 - (-3)^3}{x^5 - (-3)^5}$

Step 3: Applying the Limit Theorem

Multiply and divide the expression by $(x - (-3)) = (x + 3)$ to align with the theorem structure:

$L = \lim_{x \to -3} \left[ \frac{\frac{x^3 - (-3)^3}{x - (-3)}}{\frac{x^5 - (-3)^5}{x - (-3)}} \right]$

Applying the limit property $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$:

$L = \frac{\lim_{x \to -3} \frac{x^3 - (-3)^3}{x - (-3)}}{\lim_{x \to -3} \frac{x^5 - (-3)^5}{x - (-3)}}$

Step 4: Final Computation

By the theorem $\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$:

For the numerator ($n=3, a=-3$): $3(-3)^{3-1} = 3(-3)^2 = 3 \times 9 = 27$

For the denominator ($n=5, a=-3$): $5(-3)^{5-1} = 5(-3)^4 = 5 \times 81 = 405$

$L = \frac{27}{405}$

Simplifying the fraction by dividing both numerator and denominator by 27:

$L = \frac{1}{15}$

Visual Representation of the Function's Behavior

Final Solution: \frac{1}{15}

Theoretical Foundation: The Derivative of a Function

The derivative of a function $f(x)$ at a point $x$ is defined by the fundamental limit definition: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ [Per the First Principle of Derivatives].

Step 1: Analyzing the Standard Power Rule

For any function of the form $f(x) = x^n$, where $n$ is any real number, the derivative is calculated as: $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule of Differentiation].

Step 2: Deriving the Derivative of $x^n$ using the First Principle

To establish the identity $nx^{n-1}$, we apply the limit definition:

$f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$

Applying the Binomial Theorem for expansion: $(x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + ...$

$f'(x) = \lim_{h \to 0} \frac{(x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + ...) - x^n}{h}$

$f'(x) = \lim_{h \to 0} \frac{nx^{n-1}h + O(h^2)}{h}$

$f'(x) = \lim_{h \to 0} (nx^{n-1} + O(h)) = nx^{n-1}$

Step 3: Visual Representation of the Derivative at a Point

Step 4: Evaluating the Specific Limit Definition

The blank in the provided exercise asks for the result of the derivative process for the function $f(x) = x^n$. Based on the algebraic derivation in Step 2:

Function	Derivative (Fill in the blank)
$f(x) = x^n$	$nx^{n-1}$

Therefore, the value to be filled in the blank, representing the derivative of $x^n$ with respect to $x$, is $nx^{n-1}$.

Final Solution: The derivative of $x^n$ with respect to $x$ is $nx^{n-1}$.

Step 1: Analyzing the Limit and Indeterminacy

Given the limit: $\lim_{x \to 1} \frac{x^4 - \sqrt{x}}{x - 1}$.

First, we test for direct substitution:
Substituting $x = 1$:
Numerator: $1^4 - \sqrt{1} = 1 - 1 = 0$
Denominator: $1 - 1 = 0$
[Per the definition of indeterminate forms, $\frac{0}{0}$ requires algebraic manipulation or the application of L'Hôpital's Rule].

Step 2: Applying L'Hôpital's Rule

Since the limit results in the indeterminate form $\frac{0}{0}$, we differentiate the numerator and the denominator independently with respect to $x$ [By L'Hôpital's Rule].

Let $f(x) = x^4 - x^{1/2}$ and $g(x) = x - 1$.

Differentiating $f(x)$:
$\frac{d}{dx}(x^4) - \frac{d}{dx}(x^{1/2}) = 4x^3 - \frac{1}{2}x^{-1/2}$

Differentiating $g(x)$:
$\frac{d}{dx}(x - 1) = 1$

Step 3: Evaluating the New Limit

The limit now becomes:
$\lim_{x \to 1} \frac{4x^3 - \frac{1}{2}x^{-1/2}}{1}$

Substitute $x = 1$:
$4(1)^3 - \frac{1}{2}(1)^{-1/2}$
$= 4(1) - \frac{1}{2}(1)$
$= 4 - 0.5$
$= 3.5$

Step 4: Algebraic Verification (Rationalization Method)

Alternatively, rewrite the expression by splitting terms:
$\lim_{x \to 1} \frac{x^4 - 1 - (\sqrt{x} - 1)}{x - 1}$
$= \lim_{x \to 1} \left( \frac{x^4 - 1}{x - 1} - \frac{\sqrt{x} - 1}{x - 1} \right)$

Applying the formula $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$:
Term 1: $\lim_{x \to 1} \frac{x^4 - 1^4}{x - 1} = 4(1)^3 = 4$

Term 2: $\frac{\sqrt{x} - 1}{x - 1} = \frac{\sqrt{x} - 1}{(\sqrt{x} - 1)(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}$
$\lim_{x \to 1} \frac{1}{\sqrt{x} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$

Total Limit = $4 - \frac{1}{2} = \frac{7}{2} = 3.5$.

Final Solution: 3.5 or \frac{7}{2}

Step 1: Theoretical Foundation and Trigonometric Identities

To evaluate the limit $L = \lim_{x \to 0} \frac{1 - \cos(mx)}{1 - \cos(nx)}$, we utilize the fundamental trigonometric identity for the half-angle formula derived from the cosine double-angle identity, $\cos(2\theta) = 1 - 2\sin^2(\theta)$ [By Trigonometric Identity].

Rearranging the identity gives: $1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)$.

Step 2: Substitution into the Expression

Applying the identity to both the numerator and the denominator:

$L = \lim_{x \to 0} \frac{2\sin^2\left(\frac{mx}{2}\right)}{2\sin^2\left(\frac{nx}{2}\right)}$

Simplifying the constant factor $2$:

$L = \lim_{x \to 0} \frac{\sin^2\left(\frac{mx}{2}\right)}{\sin^2\left(\frac{nx}{2}\right)}$

Step 3: Algebraic Manipulation for Standard Limit Form

Recall the standard limit theorem: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ [By First Fundamental Limit Theorem]. To utilize this, we multiply and divide by the squared arguments of the sine functions:

$L = \lim_{x \to 0} \left[ \frac{\sin^2\left(\frac{mx}{2}\right)}{\left(\frac{mx}{2}\right)^2} \cdot \left(\frac{mx}{2}\right)^2 \right] \cdot \left[ \frac{\sin^2\left(\frac{nx}{2}\right)}{\left(\frac{nx}{2}\right)^2} \cdot \left(\frac{nx}{2}\right)^2 \right]^{-1}$

Step 4: Evaluation of Limits

Distribute the limit across the products [By Limit Laws]:

$L = \frac{\left[\lim_{x \to 0} \frac{\sin(mx/2)}{mx/2}\right]^2 \cdot \lim_{x \to 0} \left(\frac{mx}{2}\right)^2}{\left[\lim_{x \to 0} \frac{\sin(nx/2)}{nx/2}\right]^2 \cdot \lim_{x \to 0} \left(\frac{nx}{2}\right)^2}$

Since $\lim_{x \to 0} \frac{\sin(kx/2)}{kx/2} = 1$ for any constant $k \neq 0$:

$L = \frac{1^2 \cdot \left(\frac{mx}{2}\right)^2}{1^2 \cdot \left(\frac{nx}{2}\right)^2}$

Step 5: Simplification of Constants

$L = \frac{\frac{m^2 x^2}{4}}{\frac{n^2 x^2}{4}}$

Canceling the common term $\frac{x^2}{4}$ (as $x \to 0, x \neq 0$):

$L = \frac{m^2}{n^2}$

Final Solution: \frac{m^2}{n^2}

Step 1: Identifying the Indeterminate Form

Given the limit expression: $L = \lim_{x \to \frac{\pi}{4}} \frac{\tan^3(x) - \tan(\frac{\pi}{4})}{x - \frac{\pi}{4}}$

First, evaluate the expression at $x = \frac{\pi}{4}$:

• Numerator: $\tan^3(\frac{\pi}{4}) - \tan(\frac{\pi}{4}) = (1)^3 - 1 = 0$
• Denominator: $\frac{\pi}{4} - \frac{\pi}{4} = 0$

[Per the definition of indeterminate forms, $\frac{0}{0}$ necessitates the use of L'Hôpital's Rule or differentiation principles].

Step 2: Applying the Derivative Definition

Observe that the limit is in the standard form of the definition of a derivative: $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$.

Let $f(x) = \tan^3(x)$. Then, $f(\frac{\pi}{4}) = \tan^3(\frac{\pi}{4}) = 1$.

However, the numerator is $\tan^3(x) - \tan(\frac{\pi}{4})$, not $\tan^3(x) - \tan^3(\frac{\pi}{4})$. Let us re-evaluate the derivative of the specific function $g(x) = \tan^3(x)$ at $x = \frac{\pi}{4}$:

$g'(x) = \frac{d}{dx} [\tan^3(x)] = 3\tan^2(x) \cdot \sec^2(x)$ [By Chain Rule].

Step 3: Calculating the Limit using L'Hôpital's Rule

Applying L'Hôpital's Rule: $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Numerator derivative: $\frac{d}{dx} (\tan^3(x) - \tan(\frac{\pi}{4})) = 3\tan^2(x) \cdot \sec^2(x) - 0 = 3\tan^2(x)\sec^2(x)$

Denominator derivative: $\frac{d}{dx} (x - \frac{\pi}{4}) = 1$

Step 4: Substitution and Final Evaluation

Substitute $x = \frac{\pi}{4}$ into the differentiated expression:

$L = 3\tan^2(\frac{\pi}{4}) \cdot \sec^2(\frac{\pi}{4})$

Given $\tan(\frac{\pi}{4}) = 1$ and $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, hence $\sec(\frac{\pi}{4}) = \sqrt{2}$.

Square of secant: $\sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2$.

$L = 3(1)^2 \cdot (2) = 6$.

Final Solution: 6

Step 1: Establishing the First Principle Definition

To differentiate the function $f(x) = \cos(x^2 + 1)$ using the first principle, we employ the formal limit definition of the derivative: [Definition of Derivative]

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: Substitution and Expansion

Substituting $f(x) = \cos(x^2 + 1)$ and $f(x+h) = \cos((x+h)^2 + 1)$ into the formula:

$f'(x) = \lim_{h \to 0} \frac{\cos((x+h)^2 + 1) - \cos(x^2 + 1)}{h}$

Expanding the term $(x+h)^2 + 1 = x^2 + 2xh + h^2 + 1$, we have:

$f'(x) = \lim_{h \to 0} \frac{\cos(x^2 + 2xh + h^2 + 1) - \cos(x^2 + 1)}{h}$

Step 3: Applying Trigonometric Identity

We utilize the sum-to-product identity: $\cos(C) - \cos(D) = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$. [Trigonometric Transformation Identity]

Let $C = x^2 + 2xh + h^2 + 1$ and $D = x^2 + 1$.

$\frac{C+D}{2} = \frac{2x^2 + 2xh + h^2 + 2}{2} = x^2 + xh + \frac{h^2}{2} + 1$

$\frac{C-D}{2} = \frac{2xh + h^2}{2} = xh + \frac{h^2}{2}$

Step 4: Algebraic Limit Evaluation

Substitute these into the derivative expression:

$f'(x) = \lim_{h \to 0} \frac{-2 \sin(x^2 + xh + \frac{h^2}{2} + 1) \sin(xh + \frac{h^2}{2})}{h}$

To evaluate the limit, we use the standard limit property $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. [Fundamental Trigonometric Limit]

$f'(x) = \lim_{h \to 0} \left[ -\sin\left(x^2 + xh + \frac{h^2}{2} + 1\right) \cdot \frac{\sin(xh + \frac{h^2}{2})}{xh + \frac{h^2}{2}} \cdot \frac{xh + \frac{h^2}{2}}{h} \cdot 2 \right]$

As $h \to 0$:

• $\sin(x^2 + xh + \frac{h^2}{2} + 1) \to \sin(x^2 + 1)$
• $\frac{\sin(xh + \frac{h^2}{2})}{xh + \frac{h^2}{2}} \to 1$
• $\frac{2xh + h^2}{h} = 2x + h \to 2x$

Step 5: Final Computation

$f'(x) = -\sin(x^2 + 1) \cdot 1 \cdot (2x)$

$f'(x) = -2x \sin(x^2 + 1)$

Final Solution: $f'(x) = -2x \sin(x^2 + 1)$

Step 1: Identification of the Infinite Series

The given function is defined by the power series:

$y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} + \dots$

Recognizing the Taylor series expansion for the exponential function $e^x$ centered at $x=0$, we identify that: [Per the Maclaurin Series definition, $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$]

Therefore, $y = e^x$.

Step 2: Differentiation of the Series Term-by-Term

To find $\frac{dy}{dx}$, we apply the power rule for derivatives, $\frac{d}{dx}(x^n) = nx^{n-1}$, to each term of the infinite series:

$\frac{dy}{dx} = \frac{d}{dx}(1) + \frac{d}{dx}\left(\frac{x}{1!}\right) + \frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \dots$

• Term 1: $\frac{d}{dx}(1) = 0$ [Per the Constant Derivative Rule]
• Term 2: $\frac{d}{dx}\left(\frac{x}{1}\right) = 1$
• Term 3: $\frac{d}{dx}\left(\frac{x^2}{2 \cdot 1}\right) = \frac{2x}{2} = x$
• Term 4: $\frac{d}{dx}\left(\frac{x^3}{3 \cdot 2 \cdot 1}\right) = \frac{3x^2}{6} = \frac{x^2}{2!} = \frac{x^2}{2}$

Step 3: Synthesis of the Resulting Series

Substituting the derivatives back into the summation format:

$\frac{dy}{dx} = 0 + 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

Simplifying the expression:

$\frac{dy}{dx} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

Visual Representation of the Function $y=e^x$ and its Derivative

Step 4: Analytical Conclusion

Since the derived series is identical to the original series $y$, we conclude that:

$\frac{dy}{dx} = y$

Given $y = e^x$, it follows that the derivative of $e^x$ is $e^x$. [By the properties of transcendental functions, $\frac{d}{dx}(e^x) = e^x$]

Final Solution: 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...

Initial Setup & Algebraic Identification

To evaluate the limit $\lim_{x \to a} \frac{(x+2)^5 - (a+2)^5}{x-a}$, we must identify the standard algebraic form that allows for simplification. The expression matches the structure of the standard limit theorem:

$\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}$ [Per the Algebraic Limit Theorem for Powers]

Step 1: Change of Variable

To align the given expression with the standard formula, let us introduce a substitution:

Let $u = x + 2$.

As $x \to a$, it follows that $(x + 2) \to (a + 2)$. Therefore, $u \to (a+2)$.

Furthermore, since $u = x + 2$, we have $x = u - 2$ and $a = (a+2) - 2$. Subtracting these, we find:

$x - a = (u - 2) - ((a+2) - 2) = u - (a+2)$.

Step 2: Substitution into the Limit Expression

Substitute $u$ into the original limit:

$\lim_{u \to (a+2)} \frac{u^5 - (a+2)^5}{u - (a+2)}$

Step 3: Application of the Power Formula

We apply the theorem $\lim_{u \to c} \frac{u^n - c^n}{u - c} = n c^{n-1}$, where $n = 5$ and $c = (a+2)$:

$= 5 \cdot (a+2)^{5-1}$ [Applying the Binomial Limit Theorem]

$= 5(a+2)^4$

Step 4: Visualizing the Gradient

The limit represents the derivative of the function $f(x) = (x+2)^5$ evaluated at $x=a$.

Step 5: Final Derivation Summary

The derivative calculation via the limit definition gives:

$\frac{d}{dx} (x+2)^5 = 5(x+2)^{5-1} \cdot \frac{d}{dx}(x+2)$ [By Chain Rule]

$= 5(x+2)^4 \cdot 1 = 5(x+2)^4$

Evaluating at $x=a$, we obtain $5(a+2)^4$.

Final Solution: 5(a+2)^4

Step 1: Establishing the Theoretical Framework

To evaluate the limit of a trigonometric function involving $\sin$ and $\tan$ as the argument approaches zero, we employ the fundamental limit theorem: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ [By the First Fundamental Limit Theorem].

Step 2: Decomposition of the Given Expression

The expression is given as:

$\lim_{x \to 0} \frac{\sin(mx)}{\tan^3(2x)} = 2$

To isolate the standard forms, we multiply and divide by the necessary powers of the arguments:

$\lim_{x \to 0} \left[ \frac{\sin(mx)}{mx} \cdot mx \right] \cdot \left[ \frac{(2x)^3}{\tan^3(2x)} \cdot \frac{1}{(2x)^3} \right] = 2$

Step 3: Algebraic Simplification

Applying the limit properties $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$ [By Algebra of Limits]:

$\left( \lim_{x \to 0} \frac{\sin(mx)}{mx} \right) \cdot \left( \lim_{x \to 0} \left( \frac{2x}{\tan(2x)} \right)^3 \right) \cdot \lim_{x \to 0} \frac{mx}{8x^3} = 2$

Since $\lim_{x \to 0} \frac{\sin(mx)}{mx} = 1$ and $\lim_{x \to 0} \frac{2x}{\tan(2x)} = 1$, the equation reduces to:

$1 \cdot (1)^3 \cdot \lim_{x \to 0} \frac{mx}{8x^3} = 2$ $\lim_{x \to 0} \frac{m}{8x^2} = 2$

Step 4: Logical Evaluation and Constraint Analysis

Wait, we must reconsider the order of the limit. The expression $\frac{\sin(mx)}{\tan^3(2x)}$ involves $x$ in the denominator. Let us re-evaluate the expansion:

$\frac{\sin(mx)}{\tan^3(2x)} = \frac{mx + O((mx)^3)}{(2x + O((2x)^3))^3} = \frac{mx}{8x^3} = \frac{m}{8x^2}$

For the limit to exist and equal a non-zero constant ($2$), the power of $x$ in the numerator must match the power of $x$ in the denominator. If the original question implies $f(x) = \frac{\sin(mx)}{\tan^3(2x)}$, the only way to obtain a finite limit is if the expression is actually $\frac{\sin^3(mx)}{\tan^3(2x)}$ or if the denominator was $\tan(2x)$. Given the standard structure of such problems:

If we assume the intent was $\lim_{x \to 0} \frac{\sin^3(mx)}{\tan^3(2x)} = 2$:

$\lim_{x \to 0} \left( \frac{\sin(mx)}{mx} \cdot \frac{2x}{\tan(2x)} \cdot \frac{mx}{2x} \right)^3 = 2$ $(1 \cdot 1 \cdot \frac{m}{2})^3 = 2$ $\frac{m^3}{8} = 2 \implies m^3 = 16 \implies m = \sqrt[3]{16} = 2\sqrt[3]{2}$

However, strictly solving $\frac{\sin(mx)}{\tan(2x)} \cdot \frac{1}{\tan^2(2x)}$ implies that for a limit to exist as $x \to 0$, $m$ would not yield a constant. Given the standard curriculum context where the numerator power matches the denominator power:

Assuming the question implies $\lim_{x \to 0} \frac{\sin(mx)}{\tan(2x)} = 2$ (correcting for common typography in such exercises):

$\lim_{x \to 0} \frac{\sin(mx)}{mx} \cdot \frac{2x}{\tan(2x)} \cdot \frac{m}{2} = 2$ $1 \cdot 1 \cdot \frac{m}{2} = 2 \implies m = 4$

Final Solution: 4

Step 1: Identifying the Indeterminate Form

Given the limit: $\lim_{x \to \pi} \frac{1-\sin(\frac{x}{2})}{\cos(\frac{x}{2})(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))}$

Substitute $x = \pi$:

• Numerator: $1 - \sin(\frac{\pi}{2}) = 1 - 1 = 0$
• Denominator: $\cos(\frac{\pi}{2})(\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})) = 0 \cdot (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}) = 0$

[As the expression yields the form $\frac{0}{0}$, we apply trigonometric identities to simplify the limit.]

Step 2: Trigonometric Simplification

We know that $1 - \sin(\theta) = (\cos(\frac{\theta}{2}) - \sin(\frac{\theta}{2}))^2$. Setting $\theta = \frac{x}{2}$, we obtain: $1 - \sin(\frac{x}{2}) = (\cos(\frac{x}{4}) - \sin(\frac{x}{4}))^2$

For the denominator, we use the double angle identity $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$. Setting $\theta = \frac{x}{4}$: $\cos(\frac{x}{2}) = \cos^2(\frac{x}{4}) - \sin^2(\frac{x}{4}) = (\cos(\frac{x}{4}) - \sin(\frac{x}{4}))(\cos(\frac{x}{4}) + \sin(\frac{x}{4}))$

Step 3: Algebraic Substitution and Cancellation

Substitute these into the original expression: $L = \lim_{x \to \pi} \frac{(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))^2}{(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))(\cos(\frac{x}{4}) + \sin(\frac{x}{4}))(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))}$

Notice that the term $(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))$ appears in the numerator and denominator. Since $x \to \pi$, $\frac{x}{4} \to \frac{\pi}{4}$, and $\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4}) = 0$, but the terms cancel algebraically:

$L = \lim_{x \to \pi} \frac{\cos(\frac{x}{4}) - \sin(\frac{x}{4})}{(\cos(\frac{x}{4}) + \sin(\frac{x}{4}))(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))}$

Step 4: Evaluating the Limit

After canceling $(\cos(\frac{x}{4}) - \sin(\frac{x}{4}))$: $L = \lim_{x \to \pi} \frac{1}{\cos(\frac{x}{4}) + \sin(\frac{x}{4})}$

Substitute $x = \pi$: $L = \frac{1}{\cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})} = \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{1}{\frac{2}{\sqrt{2}}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

Final Solution: \frac{1}{\sqrt{2}}

Given Variables & Initial Setup

We are tasked to evaluate the limit: $$L = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos(6x)}}{\sqrt{2} \left( \frac{\pi}{3} - x \right)}$$

Step 1: Trigonometric Simplification

Utilize the half-angle identity: $1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)$. Applying this to the term $\cos(6x)$:

$$1 - \cos(6x) = 2\sin^2\left(\frac{6x}{2}\right) = 2\sin^2(3x)$$

[By trigonometric identity: $1 - \cos(2\theta) = 2\sin^2(\theta)$]

Step 2: Substitution into the Expression

Substitute the simplified numerator into the original expression:

$$L = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2\sin^2(3x)}}{\sqrt{2} \left( \frac{\pi}{3} - x \right)} = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2} \cdot |\sin(3x)|}{\sqrt{2} \left( \frac{\pi}{3} - x \right)} = \lim_{x \to \frac{\pi}{3}} \frac{|\sin(3x)|}{\frac{\pi}{3} - x}$$

Step 3: Change of Variable

Let $h = \frac{\pi}{3} - x$. As $x \to \frac{\pi}{3}$, $h \to 0$. Rearranging gives $x = \frac{\pi}{3} - h$. Substitute $x$ into the sine function:

$$\sin(3x) = \sin(3(\frac{\pi}{3} - h)) = \sin(\pi - 3h)$$

[Per the supplementary angle property: $\sin(\pi - \theta) = \sin(\theta)$]

$$\sin(\pi - 3h) = \sin(3h)$$

Step 4: Evaluating the Limit

Since we are evaluating the limit as $x \to \frac{\pi}{3}$, $3x$ approaches $\pi$. The function $\sin(3x)$ is positive for values slightly less than $\pi$ and negative for values slightly greater than $\pi$. Thus, we must evaluate the Left-Hand Limit (LHL) and Right-Hand Limit (RHL).

Case 1: $x \to \frac{\pi}{3}^-$ (which implies $h \to 0^+$)

$$L_{LHL} = \lim_{h \to 0^+} \frac{\sin(3h)}{h} = 3 \cdot \lim_{3h \to 0} \frac{\sin(3h)}{3h} = 3(1) = 3$$

[By Fundamental Limit Theorem: $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$]

Case 2: $x \to \frac{\pi}{3}^+$ (which implies $h \to 0^-$)

As $x > \frac{\pi}{3}$, $\sin(3x)$ becomes negative, so $|\sin(3x)| = -\sin(3x)$.

$$L_{RHL} = \lim_{h \to 0^-} \frac{-\sin(3h)}{h} = -3 \cdot \lim_{3h \to 0} \frac{\sin(3h)}{3h} = -3(1) = -3$$

Step 5: Conclusion on Existence

Since the Left-Hand Limit ($3$) does not equal the Right-Hand Limit ($-3$), the limit does not exist in the general sense.

Final Solution: The limit does not exist because the Left-Hand Limit is 3 and the Right-Hand Limit is -3.

Given Variables & Initial Setup

The function is defined as: $y = \dfrac{\sin(x+9)}{\cos(x)}$

We are required to find the derivative $\dfrac{dy}{dx}$ and evaluate it at $x = 0$.

Step 1: Applying the Quotient Rule

To differentiate a quotient of two functions $u(x)$ and $v(x)$, we employ the Quotient Rule [Calculus: Differentiation of Quotients]:

$\dfrac{d}{dx} \left[ \dfrac{u(x)}{v(x)} \right] = \dfrac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Let $u(x) = \sin(x+9)$ and $v(x) = \cos(x)$.

Step 2: Differentiating Individual Components

• Using the Chain Rule, $u'(x) = \dfrac{d}{dx}[\sin(x+9)] = \cos(x+9) \cdot \dfrac{d}{dx}(x+9) = \cos(x+9)$ [Chain Rule for trigonometric functions].
• $v'(x) = \dfrac{d}{dx}[\cos(x)] = -\sin(x)$ [Standard derivative of cosine].

Step 3: Constructing the Derivative Expression

Substituting the components into the quotient rule formula:

$\dfrac{dy}{dx} = \dfrac{\cos(x+9) \cdot \cos(x) - \sin(x+9) \cdot (-\sin(x))}{\cos^2(x)}$

$\dfrac{dy}{dx} = \dfrac{\cos(x+9)\cos(x) + \sin(x+9)\sin(x)}{\cos^2(x)}$

Step 4: Simplifying via Trigonometric Identity

Applying the cosine difference identity $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ [Trigonometric Addition/Subtraction Formula], where $A = x+9$ and $B = x$:

$\dfrac{dy}{dx} = \dfrac{\cos((x+9) - x)}{\cos^2(x)}$

$\dfrac{dy}{dx} = \dfrac{\cos(9)}{\cos^2(x)}$

Step 5: Evaluation at $x = 0$

Substitute $x = 0$ into the derived expression:

$\left. \dfrac{dy}{dx} \right|_{x=0} = \dfrac{\cos(9)}{\cos^2(0)}$

Since $\cos(0) = 1$, then $\cos^2(0) = 1^2 = 1$.

$\left. \dfrac{dy}{dx} \right|_{x=0} = \dfrac{\cos(9)}{1} = \cos(9)$

Visual Representation of Derivative Context

Final Solution: \cos(9)