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