Continuity practice questions

Question 1

For $f(x) = \begin{cases} x + 1, & x < 0 \ x^2 + 1, & x \ge 0 \end{cases}$, $f$ is continuous at $x = 0$ because:

Accepted Answer

$\lim_{x 	o 0^-}(x + 1) = 1$, $\lim_{x 	o 0^+}(x^2 + 1) = 1$, $f(0) = 1$. All three agree, so $f$ is continuous at $0$.

Answer

$f(0)$ is undefined

Answer

The left and right limits differ

Answer

$\lim_{x 	o 0^-}f = \lim_{x 	o 0^+}f = f(0) = 1$

Answer

$f$ is a polynomial

Question 2

The function $f(x) = x^2 + 3x - 1$ is continuous on:

Accepted Answer

Polynomial functions are continuous everywhere on $\mathbb{R}$.

Answer

$\mathbb{R} \setminus \{0\}$

Answer

$\mathbb{R} \setminus \{1\}$

Answer

$\mathbb{R}$

Answer

only on $[0, 1]$

Question 3

The function $f(x) = \dfrac{1}{x - 2}$ is discontinuous at:

Accepted Answer

$f$ is undefined at $x = 2$, where the denominator vanishes.

Answer

$0$

Answer

$2$

Answer

$-2$

Answer

nowhere

Question 4

Which of the following functions is continuous on **all** of $\mathbb{R}$?

Accepted Answer

(a) A polynomial — continuous on all of $\mathbb{R}$ ✓.

(b) Discontinuous at $x = 0$ (not defined).

(c) Discontinuous at $x = 	frac{\pi}{2} + k\pi$.

(d) Defined only for $x \ge 0$, so not continuous on negative reals.

Answer

$f(x) = x^{3} - 2x + 7$

Answer

$f(x) = \dfrac{1}{x}$

Answer

$f(x) = 	an x$

Answer

$f(x) = \sqrt{x}$

Question 5

The function $f(x) = \begin{cases} x + 1 & x < 2 \ x^{2} - 1 & x \ge 2 \end{cases}$ is continuous at $x = 2$ because:

Accepted Answer

Left limit: $\displaystyle\lim_{x 	o 2^{-}} (x + 1) = 3$.

Right limit: $\displaystyle\lim_{x 	o 2^{+}} (x^{2} - 1) = 3$.

$f(2) = 2^{2} - 1 = 3$.

All three agree, so $f$ is continuous at $x = 2$.

Answer

both one-sided limits equal $f(2)$

Answer

$f$ is differentiable at $x = 2$

Answer

$f$ is a polynomial

Answer

the limit at $x = 2$ does not exist

Question 6

If $f$ and $g$ are both continuous at $x_0$, then which is also continuous at $x_0$?

Accepted Answer

Continuity is preserved by sum, difference, product, and composition (provided $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$).

Answer

$f / g$ (always)

Answer

$|f - g|$ (only if $f(x_0) \ge g(x_0)$)

Answer

$f \circ g$ (only if $g(x_0) = x_0$)

Answer

$f + g$, $f - g$, $f \cdot g$, and $f \circ g$

Question 7

The function $f(x) = \dfrac{|x|}{x}$ has at $x = 0$:

Accepted Answer

$\lim_{x 	o 0^-}f = -1$ and $\lim_{x 	o 0^+}f = 1$ — finite but unequal limits, hence a jump discontinuity.

Answer

a removable discontinuity

Answer

a continuous extension

Answer

a jump discontinuity

Answer

an essential discontinuity

Question 8

For $f(x) = \begin{cases} 2x + 1, & x \le 1 \ x^2 + a, & x > 1 \end{cases}$, find $a$ such that $f$ is continuous at $x = 1$:

Accepted Answer

LHS: $2 \cdot 1 + 1 = 3$. RHS: $1 + a$. Setting equal: $a = 2$.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$4$

Question 9

For which value of $a \in \mathbb{R}$ is $f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1}, & x 
e 1 \ a, & x = 1 \end{cases}$ continuous at $x = 1$?

Accepted Answer

$\dfrac{x^2 - 1}{x - 1} = x + 1 	o 2$ as $x 	o 1$. For continuity, $a = 2$.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$4$

Question 10

By the Intermediate Value Theorem, the equation $f(x) = x^3 + x - 1 = 0$ has at least one root in:

Accepted Answer

$f(0) = -1 < 0$ and $f(1) = 1 > 0$. Since $f$ is continuous, by the IVT there exists $c \in (0, 1)$ with $f(c) = 0$.

Answer

$(-1, 0)$

Answer

$(0, 1)$

Answer

$(1, 2)$

Answer

nowhere on $\mathbb{R}$

Continuity

Problems & worked solutions

Problem 1 — Continuity

Problem 2 — Continuity

Problem 3 — Continuity

Problem 4 — Continuity on an Interval

Problem 5 — Continuity at a Point

Problem 6 — Continuity

Problem 7 — Continuity

Problem 8 — Continuity

Problem 9 — Continuity

Problem 10 — Continuity

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