Continuity Exercises

Approach: Check whether the function is defined at that point.

Steps:

1. The function $f(x) = \frac{1}{x}$ is undefined at $x = 0$

2. Therefore the function is discontinuous at $x = 0$

Answer: Discontinuous, because the function is undefined at $x = 0$.

Approach: Compute the left limit, the right limit and the function value separately, and see whether the three are equal.

Steps:

1. Left limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1$

2. Right limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2 \times 1 - 1 = 1$

3. Function value: $f(1) = 1^2 = 1$

4. The three are equal, so the function is continuous at $x = 1$

Answer: Continuous at $x = 1$, with no point of discontinuity.

Approach: Use the basic results about the continuity of elementary functions.

Steps:

1. $\sin x$ is a basic elementary function (a trigonometric function)

2. By the continuity theorem for elementary functions, every elementary function is continuous on its domain

3. The domain of $\sin x$ is $\mathbb{R}$

Answer: Continuous everywhere on $\mathbb{R}$.

Approach: Compute the limit and determine the type of discontinuity.

Steps:

1. The function is undefined at $x = 3$

2. Compute the limit: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 6$

3. The limit exists but is not equal to the function value (the function value is undefined)

Answer: $x = 3$ is a removable discontinuity.

Approach: Compute the left and right limits separately and compare them.

Steps:

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

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

3. Function value: $f(0) = 0 + 1 = 1$

4. The left limit is not equal to the right limit

Answer: $x = 0$ is a jump discontinuity.

Approach: Compute the left and right limits and decide.

Steps:

1. The function is undefined at $x = 0$

2. Left limit: $\lim_{x \to 0^-} \frac{1}{x^2} = +\infty$

3. Right limit: $\lim_{x \to 0^+} \frac{1}{x^2} = +\infty$

4. Neither the left nor the right limit exists (they are infinite)

Answer: $x = 0$ is a discontinuity of the second kind.

Approach: Analyse the domain of the function, then use the continuity of elementary functions.

Steps:

1. The domain of the function $f(x) = \ln(x^2 - 1)$ is given by $x^2 - 1 > 0$

2. Solve the inequality: $x^2 - 1 > 0$, giving $x < -1$ or $x > 1$

3. Therefore the domain is $(-\infty, -1) \cup (1, +\infty)$

4. By the continuity theorem for elementary functions, the function is continuous on its domain

Answer: The function is continuous on $(-\infty, -1) \cup (1, +\infty)$.

Approach: Analyse the domain of the rational function.

Steps:

1. The domain of the function $f(x) = \frac{x^2 - 4}{x - 2}$ is given by $x - 2 \neq 0$

2. That is, $x \neq 2$, so the domain is $(-\infty, 2) \cup (2, +\infty)$

3. By the continuity theorem for elementary functions, a rational function is continuous on its domain

Answer: The function is continuous on $(-\infty, 2) \cup (2, +\infty)$.

Approach: Analyse the domain of the trigonometric function.

Steps:

1. The domain of the function $f(x) = \tan x$ is given by $x \neq \frac{\pi}{2} + k\pi$ ($k$ an integer)

2. That is, the domain is $\mathbb{R} \setminus \{\frac{\pi}{2} + k\pi : k \in \mathbb{Z}\}$

3. By the continuity theorem for elementary functions, a trigonometric function is continuous on its domain

Answer: The function is continuous on $\mathbb{R} \setminus \{\frac{\pi}{2} + k\pi : k \in \mathbb{Z}\}$.

Approach: Use the zero point theorem.

Steps:

1. The function $f(x)$ is continuous on $[0, 2]$

2. $f(0) = -1 < 0$, $f(2) = 3 > 0$

3. $f(0) \cdot f(2) = (-1) \times 3 = -3 < 0$

4. By the zero point theorem, there exists $x_0 \in (0, 2)$ with $f(x_0) = 0$

Answer: There exists $x_0 \in (0, 2)$ with $f(x_0) = 0$.

Approach: Use the boundedness theorem and the extreme value theorem.

Steps:

1. By the boundedness theorem, a function continuous on a closed interval must be bounded

2. By the extreme value theorem, a function continuous on a closed interval must attain a maximum and a minimum value

3. Therefore $f(x)$ is necessarily bounded, and it attains a maximum and a minimum value

Answer: It is necessarily bounded, and it attains a maximum and a minimum value.

Approach: Use the zero point theorem.

Steps:

1. Let $f(x) = x^3 - x - 1$

2. $f(1) = 1^3 - 1 - 1 = -1 < 0$

3. $f(2) = 2^3 - 2 - 1 = 5 > 0$

4. The function is continuous on $[1, 2]$

5. By the zero point theorem, there exists $\xi \in (1, 2)$ with $f(\xi) = 0$

Answer: The equation has a solution in $(1, 2)$.

Approach: Compute the left limit, the right limit and the function value separately, and see whether the three are equal.

Steps:

1. Left limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1$

2. Right limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x - 1) = 2 \times 1 - 1 = 1$

3. Function value: $f(1) = 1^2 = 1$

4. The three are equal, so the function is continuous at $x = 1$

Answer: The function is continuous at $x = 1$.

Approach: Compute the limit and determine the type of discontinuity.

Steps:

1. The function is undefined at $x = 1$

2. Compute the limit: $\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2$

3. The limit exists but is not equal to the function value (the function value is undefined)

Answer: $x = 1$ is a removable discontinuity.

Approach: Use the zero point theorem.

Steps:

1. Let $f(x) = \cos x - x$

2. $f(0) = \cos 0 - 0 = 1 > 0$

3. $f(1) = \cos 1 - 1 < 0$ (because $\cos 1 < 1$)

4. The function is continuous on $[0, 1]$

5. By the zero point theorem, there exists $\xi \in (0, 1)$ with $f(\xi) = 0$

Answer: The equation has a solution in $(0, 1)$.

Approach: Compute the left limit, the right limit and the function value.

Steps:

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

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

3. Function value: $f(0) = 0$

4. The left limit equals the function value, but differs from the right limit

Answer: The function is discontinuous at $x = 0$; it is a jump discontinuity.

Approach: Compute the left and right limits and decide.

Steps:

1. The function is undefined at $x = 0$

2. Left limit: $\lim_{x \to 0^-} e^{\frac{1}{x}} = 0$

3. Right limit: $\lim_{x \to 0^+} e^{\frac{1}{x}} = +\infty$

4. The left and right limits are not equal, and the right limit does not exist (it is infinite)

Answer: $x = 0$ is a discontinuity of the second kind.

Approach: Use the intermediate value theorem.

Steps:

1. The function $f(x)$ is continuous on $[a, b]$

2. $f(a) = f(b)$

3. For any value between $f(a)$ and $f(b)$ (including $f(a)$ itself), the intermediate value theorem gives a $c \in (a, b)$ with $f(c) = f(a)$

Answer: There exists $c \in (a, b)$ with $f(c) = f(a)$.