Step 1: Formal Definition of Continuity

A function $f(x)$ is continuous at a point $x = c$ if and only if the following three conditions are satisfied [Per the Definition of Continuity]:

• $f(c)$ is defined.
• $\lim_{x \to c} f(x)$ exists, i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• $\lim_{x \to c} f(x) = f(c)$.

Step 2: Identifying the Domain

The given function is $f(x) = \frac{1}{x - 5}$. The function is a rational function defined for all $x \in \mathbb{R}$ except where the denominator equals zero [By the definition of the domain of rational functions]. Setting the denominator to zero: $x - 5 = 0 \implies x = 5$. Thus, the domain of $f$ is $D_f = \mathbb{R} \setminus \{5\}$.

Step 3: Analytical Verification of Continuity

Let $c$ be any arbitrary real number in the domain such that $c \neq 5$. We must evaluate $\lim_{x \to c} f(x)$:

$\lim_{x \to c} f(x) = \lim_{x \to c} \left( \frac{1}{x - 5} \right)$

Using the Quotient Law for limits, since $\lim_{x \to c} (x - 5) = c - 5 \neq 0$:

$\lim_{x \to c} f(x) = \frac{\lim_{x \to c} (1)}{\lim_{x \to c} (x - 5)} = \frac{1}{c - 5}$

Since $f(c) = \frac{1}{c - 5}$, it is evident that $\lim_{x \to c} f(x) = f(c)$. Therefore, $f(x)$ is continuous for all $c \in \mathbb{R} \setminus \{5\}$.

Step 4: Visual Representation

Step 5: Conclusion on Discontinuity

At $x = 5$, the function is undefined as division by zero is prohibited in the set of real numbers [By the Field Axioms of $\mathbb{R}$]. Because the function is not defined at $x=5$, it fails the first condition of continuity. Consequently, the function has a vertical asymptote at $x = 5$.

Final Solution: The function $f(x) = \frac{1}{x-5}$ is continuous for all $x \in \mathbb{R} \setminus \{5\}$ and is discontinuous at $x = 5$.

Step 1: Theoretical Definition of Continuity

A function $f(x)$ is continuous at a point $c$ in its domain if and only if:

1. $f(c)$ is defined.
2. $\lim_{x \to c} f(x)$ exists.
3. $\lim_{x \to c} f(x) = f(c)$.

A function is continuous on its entire domain if it is continuous at every point $c$ in its domain [Definition of Continuity on an Interval].

Step 2: Analysis of the Cosine Function: $f(x) = \cos(x)$

The domain of $f(x) = \cos(x)$ is $\mathbb{R}$. For any arbitrary point $c \in \mathbb{R}$:

Consider the limit: $\lim_{x \to c} \cos(x)$. Let $x = c + h$, where $h \to 0$.

$\lim_{h \to 0} \cos(c + h) = \lim_{h \to 0} [\cos(c)\cos(h) - \sin(c)\sin(h)]$ [Per the Trigonometric Identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$].

Since $\lim_{h \to 0} \cos(h) = 1$ and $\lim_{h \to 0} \sin(h) = 0$, we have:

$\cos(c)(1) - \sin(c)(0) = \cos(c)$.

Since $\lim_{x \to c} f(x) = f(c)$, the cosine function is continuous everywhere on $\mathbb{R}$.

Step 3: Analysis of Cosecant, Secant, and Cotangent

These functions are defined as reciprocals of sine and cosine functions. Specifically:

• $\csc(x) = \frac{1}{\sin(x)}$, defined for $\sin(x) \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$.
• $\sec(x) = \frac{1}{\cos(x)}$, defined for $\cos(x) \neq 0 \implies x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$.
• $\cot(x) = \frac{\cos(x)}{\sin(x)}$, defined for $\sin(x) \neq 0 \implies x \neq n\pi, n \in \mathbb{Z}$.

[Theorem: If $g(x)$ and $h(x)$ are continuous functions, then the quotient $f(x) = \frac{g(x)}{h(x)}$ is continuous at all points where $h(x) \neq 0$.]

Step 4: Summary of Discontinuity Points

Step 5: Visual Representation (Unit Circle-based Sine/Cosine continuity)

Final Solution: The cosine function is continuous for all $x \in \mathbb{R}$. The cosecant and cotangent functions are continuous for all $x \in \mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$. The secant function is continuous for all $x \in \mathbb{R} \setminus \{(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}\}$.

Given Variables & Initial Setup

We are tasked with examining the continuity of the function $f(x)$ defined as:

$f(x) = \frac{x^2 - 25}{x+5}, \quad x \neq -5$

To analyze the continuity at any point $c$ in the domain of the function, we must evaluate whether $\lim_{x \to c} f(x) = f(c)$ [Per the formal definition of continuity].

Step 1: Algebraic Simplification

The function is a rational function. We observe the numerator $x^2 - 25$, which is a difference of squares: $a^2 - b^2 = (a-b)(a+b)$.

$f(x) = \frac{(x-5)(x+5)}{x+5}$

Since the domain constraint is explicitly $x \neq -5$, the factor $(x+5)$ can be cancelled from the numerator and the denominator without loss of generality [By the property of limits in rational functions].

$f(x) = x - 5, \quad \forall x \in \mathbb{R} \setminus \{-5\}$

Step 2: Analysis of Continuity

A function is continuous if it is continuous at every point in its domain. Let $c$ be any real number such that $c \neq -5$.

• Calculate $f(c)$: $f(c) = c - 5$.
• Calculate the limit: $\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5$.

Since $\lim_{x \to c} f(x) = f(c)$ for all $c \in (-\infty, -5) \cup (-5, \infty)$, the function $f(x)$ is continuous at all points in its domain.

Visual Representation

Conclusion on Domain Constraint

At $x = -5$, the function $f(x)$ is undefined because the denominator becomes zero. Thus, the function is not continuous at $x = -5$ because it fails the existence condition for continuity [Per the definition of continuity at a point]. However, within the specified domain $D = \{x \in \mathbb{R} : x \neq -5\}$, the function behaves as a linear polynomial, which is continuous everywhere.

Final Solution: The function $f(x) = \frac{x^2 - 25}{x+5}$ is continuous for all $x \in \mathbb{R} \setminus \{-5\}$. It is discontinuous at $x = -5$ as the point is excluded from the domain.

Step 1: Definition of the Absolute Value Function

The function is defined as $f(x) = |x - 5|$. By the definition of the modulus function, this can be expressed as a piecewise function:

$f(x) = \begin{cases} -(x - 5) & \text{if } x < 5 \\ x - 5 & \text{if } x \ge 5 \end{cases}$

[Per the algebraic definition of absolute value $|u| = u$ if $u \ge 0$ and $-u$ if $u < 0$].

Step 2: Analysis of Continuity at an Arbitrary Point $c$

To examine the continuity of $f(x)$ for all $x \in \mathbb{R}$, we consider three distinct cases for an arbitrary real number $c$:

Case 1: $c < 5$

In the interval $(-\infty, 5)$, the function is defined by $f(x) = -x + 5$. This is a polynomial function of degree 1 [Per the theorem that every polynomial function is continuous for all real numbers].

$\lim_{x \to c} f(x) = \lim_{x \to c} (-x + 5) = -c + 5 = f(c)$. Thus, $f(x)$ is continuous for all $x < 5$.

Case 2: $c > 5$

In the interval $(5, \infty)$, the function is defined by $f(x) = x - 5$. This is also a polynomial function.

$\lim_{x \to c} f(x) = \lim_{x \to c} (x - 5) = c - 5 = f(c)$. Thus, $f(x)$ is continuous for all $x > 5$.

Case 3: $c = 5$

We must evaluate the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at $x = 5$:

• Value of function: $f(5) = |5 - 5| = 0$.
• Left-hand limit: $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} -(x - 5) = -(5 - 5) = 0$.
• Right-hand limit: $\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (x - 5) = (5 - 5) = 0$.

[Since $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$, the function is continuous at $x = 5$].

Step 3: Visual Representation

Analytical Summary

Final Solution: The function $f(x) = |x - 5|$ is continuous for all real numbers $x \in \mathbb{R}$.

Given Variables & Initial Setup

The function $f(x)$ is defined piecewise as:

For a function $f(x)$ to be continuous at a point $x = c$, the following condition must hold [Per the definition of continuity at a point]:

$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$

Step 1: Calculating the Left-Hand Limit (LHL) at $x = 5$

The LHL is evaluated using the definition of the function for $x \le 5$:

$$\text{LHL} = \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx + 1)$$

By substituting $x = 5$ into the expression [Direct Substitution Property]:

$$\text{LHL} = k(5) + 1 = 5k + 1$$

Step 2: Calculating the Right-Hand Limit (RHL) at $x = 5$

The RHL is evaluated using the definition of the function for $x > 5$:

$$\text{RHL} = \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (3x - 5)$$

By substituting $x = 5$ into the expression:

$$\text{RHL} = 3(5) - 5 = 15 - 5 = 10$$

Step 3: Calculating the Function Value at $x = 5$

The value of the function at the point of continuity is given by the definition where $x = 5$:

$$f(5) = k(5) + 1 = 5k + 1$$

Step 4: Equating Limits to Determine $k$

For $f(x)$ to be continuous at $x = 5$, the LHL must equal the RHL [Theorem: Continuity at a Point]:

$$5k + 1 = 10$$

Subtract $1$ from both sides:

$$5k = 9$$

Divide by $5$:

$$k = \frac{9}{5}$$

Visual Representation of Continuity at $x=5$

Final Solution: The value of $k$ that ensures the function $f$ is continuous at $x = 5$ is $k = \frac{9}{5}$ (or $1.8$).

Step 1: Theoretical Foundation of Continuity

A function $f(x)$ is defined to be continuous at a point $x = c$ if and only if the following condition holds [Per the definition of continuity in Calculus]:

$$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$

Given the function $f(x) = x^2 - \sin x + 5$, we must evaluate the limit as $x \to \pi$ and compare it to the function value $f(\pi)$.

Step 2: Calculating the Value of the Function at $x = \pi$

Substitute $x = \pi$ directly into the expression [By the Substitution Property of Polynomials and Trigonometric functions]:

$$f(\pi) = (\pi)^2 - \sin(\pi) + 5$$

Since $\sin(\pi) = 0$, we simplify:

$$f(\pi) = \pi^2 - 0 + 5 = \pi^2 + 5$$

Step 3: Calculating the Left-Hand and Right-Hand Limits

To find the limit $\lim_{x \to \pi} (x^2 - \sin x + 5)$, we apply the properties of limits (Sum/Difference Law) [Per the Algebra of Limits Theorem]:

$$\lim_{x \to \pi} f(x) = \lim_{x \to \pi} (x^2) - \lim_{x \to \pi} (\sin x) + \lim_{x \to \pi} (5)$$

Evaluating each term:

• $\lim_{x \to \pi} x^2 = \pi^2$
• $\lim_{x \to \pi} \sin x = \sin(\pi) = 0$
• $\lim_{x \to \pi} 5 = 5$

Combining these results:

$$\lim_{x \to \pi} f(x) = \pi^2 - 0 + 5 = \pi^2 + 5$$

Step 4: Comparison and Conclusion

We observe that:

Since the limit of the function as $x$ approaches $\pi$ is exactly equal to the value of the function at $x = \pi$, the condition for continuity is satisfied.

Final Solution: Yes, the function $f(x) = x^2 - \sin x + 5$ is continuous at $x = \pi$.

Step 1: Identifying Continuity Criteria

For a function $f(x)$ to be continuous on its entire domain $(-\infty, \infty)$, it must be continuous at every point in the domain. Specifically, it must be continuous at the transition points $x = 2$ and $x = 10$.

A function $f(x)$ is continuous at $x = c$ if and only if:

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ [Definition of continuity at a point]

Step 2: Analysis at $x = 2$

For $f(x)$ to be continuous at $x = 2$:

• Left-hand limit (LHL): $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (5) = 5$
• Right-hand limit (RHL): $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (ax + b) = 2a + b$
• Value of the function: $f(2) = 5$

Equating LHL and RHL [By definition of continuity]:

$2a + b = 5$ --- (Equation 1)

Step 3: Analysis at $x = 10$

For $f(x)$ to be continuous at $x = 10$:

• Left-hand limit (LHL): $\lim_{x \to 10^-} f(x) = \lim_{x \to 10^-} (ax + b) = 10a + b$
• Right-hand limit (RHL): $\lim_{x \to 10^+} f(x) = \lim_{x \to 10^+} (21) = 21$
• Value of the function: $f(10) = 21$

Equating LHL and RHL [By definition of continuity]:

$10a + b = 21$ --- (Equation 2)

Step 4: Solving the Simultaneous Linear Equations

We have the system of equations:

Subtract (1) from (2) [By algebraic elimination method]:

$(10a + b) - (2a + b) = 21 - 5$

$8a = 16$

$a = 2$

Substitute $a = 2$ into Equation (1):

$2(2) + b = 5$

$4 + b = 5$

$b = 1$

Step 5: Visualization of the Piecewise Function

Final Solution: The values for the constants are $a = 2$ and $b = 1$.

Step 1: Identifying the Condition for Continuity

A function $f(x)$ is continuous at a point $x = c$ if and only if:

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

In this problem, we are examining the continuity of the function $f(x)$ at the point where the definition changes, which is $x = 0$.

Step 2: Evaluating the Function at the Point

From the piecewise definition, when $x = 0$:

$f(0) = -1$ [Per the given definition $f(x) = -1$ if $x = 0$]

Step 3: Calculating the Left-Hand and Right-Hand Limits

Since the definition $f(x) = \sin x - \cos x$ holds for all $x \neq 0$, both the left-hand limit (LHL) and the right-hand limit (RHL) are calculated using this expression.

Calculating LHL:

$\lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h)$

$= \lim_{h \to 0} (\sin(0 - h) - \cos(0 - h))$

$= \lim_{h \to 0} (-\sin h - \cos h)$ [Since $\sin(-h) = -\sin h$ and $\cos(-h) = \cos h$]

$= -\sin(0) - \cos(0) = 0 - 1 = -1$

Calculating RHL:

$\lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h)$

$= \lim_{h \to 0} (\sin(0 + h) - \cos(0 + h))$

$= \lim_{h \to 0} (\sin h - \cos h)$

$= \sin(0) - \cos(0) = 0 - 1 = -1$

Step 4: Comparison and Conclusion

Comparing the values obtained:

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1$, the function satisfies the criteria for continuity at $x = 0$ [By the definition of continuity of a function at a point]. For all other $x \neq 0$, the function $f(x) = \sin x - \cos x$ is continuous as it is the difference of two continuous trigonometric functions.

Final Solution: The function $f(x)$ is continuous for all $x \in \mathbb{R}$.

Theoretical Foundation

A function $f(x)$ is defined to be continuous at a point $x = c$ if and only if the following three conditions are satisfied:

• $f(c)$ is defined.
• $\lim_{x \to c} f(x)$ exists, i.e., $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.
• $\lim_{x \to c} f(x) = f(c)$.

Step 1: Analysis at $x = 0$

Given $f(x) = 5x - 3$:

• Value of the function at $x = 0$: $f(0) = 5(0) - 3 = -3$.
• Limit at $x = 0$: $\lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$. [By Algebraic Limit Theorem].
• Since $\lim_{x \to 0} f(x) = f(0) = -3$, the function $f(x)$ is continuous at $x = 0$.

Step 2: Analysis at $x = -3$

• Value of the function at $x = -3$: $f(-3) = 5(-3) - 3 = -15 - 3 = -18$.
• Limit at $x = -3$: $\lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -18$.
• Since $\lim_{x \to -3} f(x) = f(-3) = -18$, the function $f(x)$ is continuous at $x = -3$.

Step 3: Analysis at $x = 5$

• Value of the function at $x = 5$: $f(5) = 5(5) - 3 = 25 - 3 = 22$.
• Limit at $x = 5$: $\lim_{x \to 5} (5x - 3) = 5(5) - 3 = 22$.
• Since $\lim_{x \to 5} f(x) = f(5) = 22$, the function $f(x)$ is continuous at $x = 5$.

Visual Representation of $f(x) = 5x - 3$

Conclusion

As demonstrated across all three test points, the limit of the function equals the value of the function, confirming polynomial continuity [By the Theorem that all polynomial functions are continuous for all $x \in \mathbb{R}$].

Final Solution: The function $f(x) = 5x - 3$ is continuous at $x = 0$, $x = -3$, and $x = 5$ as $\lim_{x \to c} f(x) = f(c)$ for all $c \in \{0, -3, 5\}$.

Step 1: Establishing the Domain and Point of Discontinuity

The function $f(x)$ is defined piecewise. For $x \neq 0$, the function is defined as $f(x) = x^2 \sin\left(\frac{1}{x}\right)$. At $x = 0$, the function is defined as $f(0) = 0$. Since the transition between these two definitions occurs at $x = 0$, we must test the continuity specifically at $x = 0$. [Per the definition of continuity, $f(x)$ is continuous at $x = c$ if $\lim_{x \to c} f(x) = f(c)$].

Step 2: Evaluating the Limit at $x = 0$

To determine if the limit exists as $x \to 0$, we evaluate:

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

We apply the Squeeze Theorem (also known as the Sandwich Theorem). We know the range of the sine function is bounded:

$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1, \quad \forall x \neq 0$

Step 3: Applying Algebraic Bounds

Multiplying the inequality by $x^2$ (which is always positive for $x \neq 0$):

$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$

Now, we evaluate the limits of the bounding functions:

• $\lim_{x \to 0} (-x^2) = 0$
• $\lim_{x \to 0} (x^2) = 0$

[By the Squeeze Theorem, if $g(x) \leq f(x) \leq h(x)$ and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$].

Therefore, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

Step 4: Comparison with Function Value

We have calculated:

$\lim_{x \to 0} f(x) = 0$

From the problem definition:

$f(0) = 0$

Since $\lim_{x \to 0} f(x) = f(0) = 0$, the function satisfies the condition for continuity at $x = 0$.

Step 5: Visual Representation of Bounding

Final Solution: The function $f(x)$ is continuous at $x = 0$, and since it is composed of continuous functions for all $x \neq 0$, the function $f(x)$ is continuous for all $x \in \mathbb{R}$.

Step 1: Analyzing the Domain and Potential Points of Discontinuity

The function $f(x)$ is defined piecewise over the set of all real numbers $\mathbb{R}$. The transition point for the definition is at $x = 1$.

• Case 1: $x < 1$, $f(x) = x^2 + 1$ [A polynomial function, which is continuous for all $x \in (-\infty, 1)$].
• Case 2: $x > 1$, $f(x) = x + 1$ [A polynomial function, which is continuous for all $x \in (1, \infty)$].
• Case 3: $x = 1$, the point of interest where the definition changes.

Step 2: Checking Continuity at $x = 1$

For $f(x)$ to be continuous at $x = c$, the following condition must hold: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.

Left-Hand Limit (LHL):

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

$\lim_{x \to 1^-} f(x) = (1)^2 + 1 = 2$ [By direct substitution into the polynomial $x^2 + 1$].

Right-Hand Limit (RHL):

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1)$

$\lim_{x \to 1^+} f(x) = 1 + 1 = 2$ [By direct substitution into the polynomial $x + 1$].

Value of Function at $x = 1$:

$f(1) = 1 + 1 = 2$ [Using the definition $f(x) = x + 1$ for $x \ge 1$].

Step 3: Verification of Limit Equality

Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 2$, the function satisfies the criteria for continuity at $x = 1$.

Step 4: Final Assessment

There are no values of $x$ for which the function is discontinuous.

Final Solution: The function $f$ has no points of discontinuity.

Step 1: Decomposition into Composite Functions

To examine the continuity of the function $f(x) = \sin|x|$, we represent it as a composite function $f(x) = (g \circ h)(x)$, where:

• $h(x) = |x|$
• $g(x) = \sin(x)$

By the definition of composition, $(g \circ h)(x) = g(h(x)) = \sin(|x|)$. [Per the definition of composite functions].

Step 2: Analysis of Component Functions

We analyze the continuity of the individual components over the entire real domain, $\mathbb{R}$:

• For $h(x) = |x|$: The absolute value function is known to be continuous for all $x \in \mathbb{R}$. Formally, $\lim_{x \to a} |x| = |a|$ for any $a \in \mathbb{R}$. [By the property of modulus functions].
• For $g(x) = \sin(x)$: The sine function is a standard trigonometric function known to be continuous for all $x \in \mathbb{R}$. [By the fundamental properties of trigonometric functions].

Step 3: Application of the Composition Theorem

The Continuity Theorem states: If $h(x)$ is continuous at $x = a$ and $g(x)$ is continuous at $h(a)$, then the composite function $(g \circ h)(x)$ is continuous at $x = a$.

1. Let $a$ be any real number.
2. $h(x) = |x|$ is continuous at $x = a$.
3. Let $b = h(a) = |a|$. Since $g(x) = \sin(x)$ is continuous for all real numbers, it is necessarily continuous at $b = |a|$.
4. Since both conditions are satisfied for all $a \in \mathbb{R}$, $f(x) = g(h(x))$ is continuous on its entire domain. [By the Theorem of Continuity of Composite Functions].

Visual Representation

Step 4: Algebraic Verification at Critical Point $x=0$

To ensure rigor, we verify the limit at $x=0$:

$\lim_{x \to 0^-} \sin|x| = \lim_{x \to 0^-} \sin(-x) = \sin(0) = 0$

$\lim_{x \to 0^+} \sin|x| = \lim_{x \to 0^+} \sin(x) = \sin(0) = 0$

$f(0) = \sin|0| = 0$

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$, the function is continuous at the origin. [By the definition of limit existence and continuity].

Final Solution: The function $f(x) = \sin|x|$ is continuous for all $x \in \mathbb{R}$ because it is the composition of two continuous functions, $\sin(x)$ and $|x|$, which are continuous over their entire domain.

Step 1: Theoretical Requirement for Continuity

A function $f(x)$ is defined as continuous at a point $x = c$ if and only if the left-hand limit, the right-hand limit, and the value of the function at that point are equal [Per the definition of continuity at a point]:

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

In this specific case, $c = 3$.

Step 2: Evaluating the Left-Hand Limit (LHL)

The LHL is evaluated as $x$ approaches 3 from the left side ($x < 3$). In this domain, the function is defined as $f(x) = ax + 1$.

$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (ax + 1)$

By direct substitution [Limit Law for Polynomials]:

$\text{LHL} = a(3) + 1 = 3a + 1$

Step 3: Evaluating the Right-Hand Limit (RHL)

The RHL is evaluated as $x$ approaches 3 from the right side ($x > 3$). In this domain, the function is defined as $f(x) = bx + 3$.

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

By direct substitution:

$\text{RHL} = b(3) + 3 = 3b + 3$

Step 4: Evaluating the Value of the Function at $x = 3$

For $x = 3$, the function definition includes the equality $x \le 3$, thus $f(x) = ax + 1$.

$f(3) = a(3) + 1 = 3a + 1$

Step 5: Establishing the Algebraic Relationship

For the function to be continuous at $x = 3$, we equate the LHL and the RHL:

$3a + 1 = 3b + 3$

Isolating the variables to define the relationship:

$3a - 3b = 3 - 1$

$3(a - b) = 2$

$a - b = \frac{2}{3}$

Or, expressed in terms of $a$:

$a = b + \frac{2}{3}$

Visual Representation of Continuity

Final Solution: The relationship required for the function to be continuous at $x = 3$ is $3a - 3b = 2$, or $a = b + \frac{2}{3}$.

Step 1: Theoretical Analysis of the Function

The function $f(x)$ is defined piecewise as follows:

$f(x) = \begin{cases} \frac{x}{|x|}, & x < 0 \\ -1, & x \ge 0 \end{cases}$

For $x < 0$, $|x| = -x$. Therefore, the function simplifies to:

$f(x) = \frac{x}{-x} = -1$ for all $x < 0$.

Thus, the function can be expressed consistently as:

$f(x) = -1$ for all $x \in \mathbb{R}$.

Step 2: Checking Continuity at General Points

To determine points of discontinuity, we examine the limit of the function as $x \to c$ for any real number $c$.

• Case 1: $c < 0$
  $\lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1$. Since $f(c) = -1$, the function is continuous for all $c < 0$.
• Case 2: $c > 0$
  $\lim_{x \to c} f(x) = \lim_{x \to c} (-1) = -1$. Since $f(c) = -1$, the function is continuous for all $c > 0$.

Step 3: Checking Continuity at the Critical Point ($x = 0$)

We evaluate the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the value of the function at $x = 0$.

LHL: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( \frac{x}{|x|} \right) = \lim_{x \to 0^-} \left( \frac{x}{-x} \right) = -1$. [By definition of absolute value for negative reals].

RHL: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1$. [Per the given definition for $x \ge 0$].

Value of function: $f(0) = -1$. [Per the given definition $f(x) = -1$ when $x = 0$].

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = -1$, the function satisfies the condition for continuity at $x = 0$.

Step 4: Visualizing the Function

Summary of Analytical Results

Final Solution: The function $f(x) = -1$ for all $x \in \mathbb{R}$ is a constant function; therefore, there are no points of discontinuity for the given function.

Step 1: Decomposition into Composite Functions

To analyze the continuity of the function $f(x) = |\cos x|$, we express $f(x)$ as a composite of two simpler functions. Let:

• $g(x) = \cos x$
• $h(x) = |x|$

Then, the composite function $(h \circ g)(x) = h(g(x)) = |\cos x| = f(x)$. [Per the definition of composite functions: $(h \circ g)(x) = h(g(x))]$.

Step 2: Analysis of Component Functions

We evaluate the continuity of the individual components over the domain of real numbers, $x \in \mathbb{R}$:

1. Continuity of $g(x) = \cos x$: The trigonometric function $g(x) = \cos x$ is a standard transcendental function known to be continuous for all $x \in \mathbb{R}$. [By the Property of Trigonometric Continuity].
2. Continuity of $h(x) = |x|$: The absolute value (modulus) function is continuous for all $x \in \mathbb{R}$. Its limit at any point $c$ is $|c|$, which equals $h(c)$. [By the Limit definition of Modulus Functions].

Step 3: Application of the Composition Theorem

The Continuity Theorem states that if a function $g$ is continuous at a point $c$, and a function $h$ is continuous at $g(c)$, then the composite function $(h \circ g)$ is continuous at $c$. [Per the Composition of Continuous Functions Theorem].

Since $g(x) = \cos x$ is continuous for all $x \in \mathbb{R}$ and $h(x) = |x|$ is continuous for all $x \in \mathbb{R}$, their composition $f(x) = h(g(x)) = |\cos x|$ must be continuous for all $x \in \mathbb{R}$.

Step 4: Graphical Representation of $|\cos x|$

Step 5: Rigorous Verification via Limits

For any arbitrary point $c \in \mathbb{R}$:

$\lim_{x \to c} f(x) = \lim_{x \to c} |\cos x|$

Since the absolute value function is continuous, we can pull the limit inside the modulus operator [By the continuity property of the modulus function]:

$\lim_{x \to c} |\cos x| = |\lim_{x \to c} \cos x| = |\cos c| = f(c)$

Since $\lim_{x \to c} f(x) = f(c)$ for all $c \in \mathbb{R}$, the function is continuous everywhere in its domain.

Final Solution: The function $f(x) = |\cos x|$ is a composite of two continuous functions, $\cos x$ and $|x|$, and therefore satisfies the condition $\lim_{x \to c} f(x) = f(c)$ for all $x \in \mathbb{R}$, proving it is a continuous function.

Step 1: Formal Definition of Continuity

A function $f(x)$ is defined as continuous at a point $x = c$ if and only if the following three conditions are satisfied [Per the Limit Definition of Continuity]:

• The function $f(c)$ is defined.
• The limit $\displaystyle \lim_{x \to c} f(x)$ exists.
• The limit is equal to the functional value: $\displaystyle \lim_{x \to c} f(x) = f(c)$.

Step 2: Evaluating the Functional Value at $x = 3$

Given the function $f(x) = 2x^2 - 1$, we substitute $x = 3$ to find the value of the function at the point of interest:

$f(3) = 2(3)^2 - 1$

$f(3) = 2(9) - 1$

$f(3) = 18 - 1 = 17$

[By the Principle of Substitution for Polynomials]

Step 3: Calculating the Left-Hand and Right-Hand Limits

We evaluate the limit of the function as $x$ approaches 3 from both sides:

Left-Hand Limit (LHL): $\displaystyle \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h)$

$\displaystyle \lim_{h \to 0} [2(3 - h)^2 - 1] = 2(3 - 0)^2 - 1 = 17$

Right-Hand Limit (RHL): $\displaystyle \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3 + h)$

$\displaystyle \lim_{h \to 0} [2(3 + h)^2 - 1] = 2(3 + 0)^2 - 1 = 17$

[Since LHL = RHL, $\displaystyle \lim_{x \to 3} f(x) = 17$]

Step 4: Comparative Analysis

Step 5: Visual Representation of the Parabolic Path

Conclusion

Since $\displaystyle \lim_{x \to 3} f(x) = f(3) = 17$, the conditions for continuity are satisfied [By the Algebraic Property of Continuous Functions, all polynomial functions are continuous for all $x \in \mathbb{R}$].

Final Solution: The function $f(x) = 2x^2 - 1$ is continuous at $x = 3$.

Step 1: Identifying the Critical Point

The function $f(x)$ is defined as piecewise across the domain $\mathbb{R}$. The transition point where the algebraic definition changes is $x = 2$. For any $x < 2$, the function is a polynomial $x^3 - 3$, and for $x > 2$, it is a polynomial $x^2 + 1$. Since polynomials are continuous everywhere in their domain [By the Algebra of Continuous Functions], we need only examine the behavior of $f(x)$ at the critical point $x = 2$.

Step 2: Evaluating the Left-Hand Limit (LHL)

The LHL is calculated as $x$ approaches 2 from the negative direction ($x \to 2^-$):

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

Substituting $x = 2$:

$LHL = (2)^3 - 3 = 8 - 3 = 5$

Step 3: Evaluating the Right-Hand Limit (RHL)

The RHL is calculated as $x$ approaches 2 from the positive direction ($x \to 2^+$):

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

Substituting $x = 2$:

$RHL = (2)^2 + 1 = 4 + 1 = 5$

Step 4: Evaluating the Functional Value at $x=2$

Per the definition $f(x) = x^3 - 3$ when $x \le 2$:

$f(2) = (2)^3 - 3 = 8 - 3 = 5$

Step 5: Synthesizing Continuity at $x=2$

A function $f(x)$ is continuous at $x = c$ if and only if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ [Definition of Continuity].

From our calculations:

• $LHL = 5$
• $RHL = 5$
• $f(2) = 5$

Since $LHL = RHL = f(2)$, the function is continuous at $x = 2$.

Final Solution: The function $f(x)$ is continuous for all $x \in \mathbb{R}$. There are no points of discontinuity.

Step 1: Analyzing Continuity at $x = 0$

For a function $f(x)$ to be continuous at a point $x = c$, the following condition must hold: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$ [Per the definition of continuity].

Given the function: $f(x) = \begin{cases} \lambda(x^2 - 2x), & x \le 0 \\ 4x + 1, & x > 0 \end{cases}$

Left-Hand Limit (LHL) at $x = 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = 0$

Right-Hand Limit (RHL) at $x = 0$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 4(0) + 1 = 1$

Value at $x = 0$: $f(0) = \lambda(0^2 - 2(0)) = 0$

For continuity at $x = 0$, we require $\text{LHL} = \text{RHL} = f(0)$. Substituting the values obtained: $0 = 1 = 0$. Since $0 \neq 1$, the function is discontinuous at $x = 0$ for all real values of $\lambda$. No value of $\lambda$ can satisfy this equality.

Step 2: Analyzing Continuity at $x = 1$

At $x = 1$, we observe the interval $x > 0$. The function is defined as $f(x) = 4x + 1$ for all values in the neighborhood of $x = 1$ (specifically, in the interval $(0, \infty)$).

Limit at $x = 1$: $\lim_{x \to 1} f(x) = \lim_{x \to 1} (4x + 1) = 4(1) + 1 = 5$

Value at $x = 1$: $f(1) = 4(1) + 1 = 5$ [Since $1 > 0$, we use the second branch of the function].

Since $\lim_{x \to 1} f(x) = f(1) = 5$, the function $f(x)$ is continuous at $x = 1$ for any value of $\lambda$.

Step 3: Visual Representation

Final Solution: There is no value of $\lambda$ for which the function is continuous at $x = 0$. The function is continuous at $x = 1$ for all real values of $\lambda$.

Initial Definition and Theoretical Foundation

To demonstrate that the function $f(x) = \cos(x^2)$ is continuous for all $x \in \mathbb{R}$, we utilize the Theorem on the Continuity of Composite Functions. This theorem states: If a function $g(x)$ is continuous at $c$ and a function $h(x)$ is continuous at $g(c)$, then the composite function $(h \circ g)(x) = h(g(x))$ is continuous at $c$.

Step 1: Decomposition of the Function

We express $f(x)$ as the composition of two simpler functions, $g(x)$ and $h(x)$, such that:

• Let $g(x) = x^2$
• Let $h(x) = \cos(x)$
• Then, $(h \circ g)(x) = h(g(x)) = \cos(x^2) = f(x)$

Step 2: Proving Continuity of Constituent Functions

Part A: Continuity of $g(x)$

The function $g(x) = x^2$ is a polynomial function of degree 2. [By the property that all polynomial functions are continuous for every real number $x \in \mathbb{R}$], we conclude that $g(x)$ is continuous everywhere.

Part B: Continuity of $h(x)$

The function $h(x) = \cos(x)$ is a trigonometric function. [Per the standard trigonometric limit theorem, $\lim_{x \to c} \cos(x) = \cos(c)$ for any $c \in \mathbb{R}$]. Thus, $h(x)$ is continuous everywhere on its domain, $\mathbb{R}$.

Step 3: Synthesis of the Composite Function

Since $g(x)$ is continuous at all $x \in \mathbb{R}$ and $h(x)$ is continuous at all points in its domain (which includes the range of $g(x)$), the composite function $f(x) = h(g(x))$ must be continuous for all $x \in \mathbb{R}$. [By the Theorem of Continuity of Composite Functions].

Graphical Representation

The graph demonstrates that the function oscillates with increasing frequency as $|x|$ increases, maintaining a smooth, unbroken path across the entire domain, confirming continuity.

Final Solution: Since $g(x) = x^2$ and $h(x) = \cos(x)$ are both continuous functions on $\mathbb{R}$, their composition $f(x) = (h \circ g)(x) = \cos(x^2)$ is continuous for all $x \in \mathbb{R}$.

Step 1: Theoretical Definition of the Function

The function $f(x)$ is defined piecewise based on the absolute value of $x$. We must express the modulus function $|x|$ according to its definition: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

Substituting this into the given definition:

$f(x) = \begin{cases} \frac{-x}{x} = -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ \frac{x}{x} = 1, & \text{if } x > 0 \end{cases}$

Simplifying further:

$f(x) = \begin{cases} -1, & \text{if } x < 0 \\ 0, & \text{if } x = 0 \\ 1, & \text{if } x > 0 \end{cases}$

Step 2: Checking Continuity for $x < 0$ and $x > 0$

For any real number $c < 0$, the function is a constant function $f(x) = -1$. Since all constant functions are continuous on their domain [Per the Property of Constant Functions], $f(x)$ is continuous for all $x < 0$.

For any real number $c > 0$, the function is a constant function $f(x) = 1$. Similarly, $f(x)$ is continuous for all $x > 0$.

Step 3: Checking Continuity at $x = 0$

A function $f(x)$ is continuous at $x = c$ if and only if $\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. [Per the Definition of Continuity].

• Left-Hand Limit (LHL): $\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-1) = -1$
• Right-Hand Limit (RHL): $\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (1) = 1$
• Function Value: $f(0) = 0$

Since $\displaystyle \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$, the limit $\displaystyle \lim_{x \to 0} f(x)$ does not exist. Therefore, the function has a jump discontinuity at $x = 0$.

Step 4: Visual Representation

Step 5: Tabular Summary

Final Solution: The function $f(x)$ is continuous for all $x \in \mathbb{R} \setminus \{0\}$. The only point of discontinuity is $x = 0$.