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📝 Continuity of composite functions (15 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 15 questions available

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What is Continuity of composite functions?

Definition:
If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then the composite function $f \circ g$ (i.e., $f(g(x))$ ) is continuous at $a$ . This means that the limit of a composite can be found by applying the outer function to the limit of the inner, $\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$ , provided the inner limit exists and the outer is continuous at that value.

Example:
Show $h(x) = \sin(x^2)$ is continuous at $x=0$ .
Solution: $g(x)=x^2$ continuous, $f(u)=\sin u$ continuous, so $h$ continuous at $0$ ; limit $= \sin(0)=0$ .

Reason:
This theorem allows us to extend continuity to complex functions, which is essential for evaluating limits of nested functions and for ensuring that compositions of well-behaved functions remain well-behaved in applications.

Easy

Medium

Hard

📝 All Continuity of composite functions MCQs

Q1. What does it mean for a function $f$ to be continuous at a point $a$ ?

A.The limit

\lim_{x\to a} f(x)

exists and equals

f(a)

. ✅

B.The function

f

is differentiable at

a

C.Both the limit exists and equals

f(a)

D.Only that

f(a)

is defined.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A function is continuous at $a$ precisely when the limit as $x$ approaches $a$ exists and matches the function’s value at that point. This captures the intuitive idea that there is no jump or break in the graph at $a$ .

Q2. According to Theorem 1.5.5, if $\lim_{x\to c} g(x)=L$ and $f$ is continuous at $L$ , then which of the following equalities holds?

\lim_{x\to c} f(g(x)) = L

\lim_{x\to c} f(g(x)) = f(L)

\lim_{x\to c} g(f(x)) = f(L)

. ✅

\lim_{x\to c} f(g(x))

does not exist.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The theorem explicitly allows the limit operator to pass through a continuous outer function, yielding $\lim_{x\to c} f(g(x)) = f\big(\lim_{x\to c} g(x)\big)=f(L)$ . The other options either misuse composition order or ignore the continuity condition.

Q3. If $g$ is continuous at $c$ but $f$ is discontinuous at $g(c)$ , what can be concluded about the continuity of the composition $f\circ g$ at $c$ ?

A.It must be continuous because

g

is continuous.

B.It is discontinuous because the discontinuity of

f

is transferred through the composition.

C.Continuity cannot be determined without more information about

f

. ✅

D.It is continuous only from the left.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Continuity of a composition requires the outer function to be continuous at the image point. Since $f$ fails that condition at $g(c)$ , the composition inherits the discontinuity, making $f\circ g$ discontinuous at $c$ .

Q4. Suppose $\lim_{x\to c^-} g(x)=L$ and $f$ is continuous from the right at $L$ . Which of the following statements is true about $\lim_{x\to c^-} f(g(x))$ ?

A.The limit equals

f(L)

. ✅

B.The limit does not exist.

C.The limit equals

L

D.Continuity of

f

from the right is irrelevant.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: When the inner limit approaches $L$ from the left and the outer function is continuous from the right at that same point, the composition’s left‑hand limit equals the outer function evaluated at $L$ , i.e., $f(L)$ . The directional continuity of $f$ is sufficient.

Q5. Let $g(x)=\begin{cases} x^2 & x\le 0\\ 2x & x>0 \end{cases}$ and $f(y)=\sqrt{y+1}$ . Determine whether $\lim_{x\to 0} f(g(x))$ exists and, if so, its value.

A.The limit exists and equals

1

B.The limit exists and equals

\sqrt{1}

C.The limit does not exist because

g

is not continuous at 0.

D.The limit exists and equals

\sqrt{0+1}=1

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Both one‑sided definitions of $g$ approach 0 as $x\to0$ , so $\lim_{x\to0}g(x)=0$ . Since $f(y)=\sqrt{y+1}$ is continuous at $y=0$ , the composition limit equals $f(0)=\sqrt{1}=1$ .

Q6. Compare the continuity of the composition $|g(x)|$ when $g$ is continuous at $c$ versus when $g$ has a removable discontinuity at $c$ .

A.Both are continuous at

c

because absolute value preserves continuity.

B.If

g

has a removable discontinuity,

|g(x)|

may become continuous after taking absolute value. ✅

C.If

g

is continuous at

c

|g(x)|

is always discontinuous at

c

D.The continuity of

|g(x)|

is unrelated to the continuity of

g

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A removable jump where $g$ changes sign can be eliminated by the absolute value, turning a pointwise break into a continuous value. When $g$ is already continuous, $|g|$ remains continuous, but the absolute value can also repair certain sign‑related gaps.

Q7. Given two functions $f$ and $g$ where $f$ is continuous everywhere and $g$ is continuous at $c$ , which of the following statements best describes the continuity of $f\circ g$ at $c$ compared to $g$ alone?

A.The composition inherits any discontinuities of

g

c

B.The composition is always continuous at

c

regardless of

g

C.The composition is continuous at

c

only if

g(c)

lies in a region where

f

is differentiable.

D.The composition is continuous at

c

if and only if

g

is continuous at

c

. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Since $f$ is continuous everywhere, the only obstacle to continuity of the composition is the continuity of the inner function. Thus $f\circ g$ is continuous at $c$ exactly when $g$ is continuous there, matching the theorem’s condition.

Q8. Let $f(x)=\begin{cases}x^2 & x\le 1\\ 2x-1 & x>1\end{cases}$ and $g(x)=\sqrt{x}$ . Analyze the continuity of the composition $h(x)=f(g(x))$ at $x=1$ .

A.h is continuous at 1 because both f and g are continuous at the relevant points. ✅

B.h is discontinuous at 1 because f is not continuous at g(1).

C.h is continuous at 1 but not differentiable there.

D.h is undefined at 1.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At $x=1$ , $g(1)=1$ . Both pieces of $f$ meet at $x=1$ with the same value (1) and matching limits, so $f$ is continuous there. Since $g$ is continuous everywhere, the composition inherits continuity, giving $h$ continuity at $x=1$ .

Q9. If $g$ is continuous at $c$ and $f$ is the absolute value function, what can be said about the limit $\lim_{x\to c}|g(x)|$ ?

A.It equals

|g(c)|

B.It does not exist unless

g(c)=0

C.It equals

\lim_{x\to c} g(x)

. ✅

D.It equals

|\lim_{x\to c} g(x)|

only if the limit exists.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The absolute value function is continuous everywhere, so the limit of $|g(x)|$ as $x$ approaches $c$ is simply the absolute value of the limit of $g(x)$ , which equals $|g(c)|$ because $g$ itself is continuous at $c$ .

Q10. Consider the composition $h(x)=\ln(\sqrt{x^2+1})$ . Using the continuity of the inner functions, determine whether $h$ is continuous for all real $x$ .

A.Yes, because both

\sqrt{x^2+1}

and

\ln

are continuous on their domains, and the inner expression is always positive. ✅

B.No, because

\ln

is not defined for non‑positive arguments.

C.It is continuous only for

x\neq 0

D.Continuity cannot be concluded without evaluating limits.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The inner function $\sqrt{x^2+1}$ yields values $\ge 1$ for every real $x$ , keeping the argument of $\ln$ strictly positive. Both the square‑root and natural‑log functions are continuous on their respective domains, so their composition is continuous for all real numbers.

Q11. Let $g(x)=\frac{x^2-4}{x-2}$ for $x\neq 2$ and define $g(2)=4$ . Let $f(y)=|y|$ . Determine whether the composition $f\circ g$ is continuous at $x=2$ .

A.It is continuous because the limit of

g(x)

x\to2

exists and equals 4, and

|y|

is continuous. ✅

B.It is discontinuous because

g

has a removable discontinuity at 2.

C.It is continuous only from the left.

D.It is discontinuous because the absolute value function introduces a cusp.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The simplified form of $g(x)$ is $x+2$ for $x\neq2$ , whose limit as $x\to2$ is 4. By defining $g(2)=4$ , $g$ becomes continuous at 2. Since the absolute value function is continuous everywhere, the composition inherits this continuity, making it continuous at $x=2$ .

Q12. Suppose $g$ is continuous at $c$ and $f$ is continuous everywhere except at a single point $p\neq g(c)$ . What can be said about the continuity of $f\circ g$ at $c$ ?

A.It is continuous at

c

because the discontinuity of

f

does not affect the composition.

B.It is discontinuous at

c

because any discontinuity in

f

propagates through the composition.

C.Continuity at

c

depends on whether

g(c)=p

. ✅

D.The composition is continuous only if

g

is differentiable at

c

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Since $f$ is continuous at every point except $p$ and $g(c)\neq p$ , the outer function is continuous at the specific argument $g(c)$ . The inner function’s continuity guarantees that the composition meets the theorem’s conditions, ensuring continuity at $c$ .

Q13. If a function $g$ is continuous on $[a,b]$ and $f$ is continuous on the range of $g$ , which theorem guarantees that $f\circ g$ attains its maximum on $[a,b]$ ?

A.Intermediate Value Theorem.

B.Extreme Value Theorem applied to the composition. ✅

C.Mean Value Theorem.

D.Bolzano's Theorem.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The Extreme Value Theorem states that a continuous function on a closed interval attains both a maximum and a minimum. Since $g$ maps $[a,b]$ continuously into a compact range and $f$ is continuous on that range, the composition $f\circ g$ is continuous on $[a,b]$ and thus also satisfies the Extreme Value Theorem.

Q14. Given that $\lim_{x\to c^+} g(x)=L$ and $f$ is continuous at $L$ but not defined for arguments less than $L$ , what can be said about $\lim_{x\to c^+} f(g(x))$ ?

A.The limit exists and equals

f(L)

. ✅

B.The limit does not exist because

f

is undefined for some values.

C.The limit equals

L

D.Continuity of

f

from the right is insufficient to determine the limit.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Even though $f$ lacks a definition for inputs below $L$ , the right‑hand limit of $g$ approaches $L$ from above, staying within the domain where $f$ is defined. The continuity of $f$ at $L$ then ensures the composition’s right‑hand limit equals $f(L)$ .

Q15. Let $g(x)=\sin\frac{1}{x}$ for $x\neq 0$ and $g(0)=0$ . Let $f(y)=y^2$ . Analyze the continuity of $f\circ g$ at $x=0$ .

A.Continuous because

f

smooths out the oscillations of

g

B.Discontinuous because

g

is not continuous at 0. ✅

C.Continuous only from the right.

D.Undefined at

x=0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The inner function $g$ oscillates without settling as $x\to0$ , so its limit does not exist. Squaring does not eliminate this oscillation; $(\sin(1/x))^2$ still varies between 0 and 1. Consequently, the composition fails to have a limit at 0 and is discontinuous there.

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