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📝 Limits of piecewise functions (14 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available

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What is Limits of piecewise functions?

Definition:
For a piecewise function defined by different expressions on different intervals, the limit at a boundary point $a$ exists only if the left-hand and right-hand limits, computed using the respective sub-function for each side, are equal. This requires evaluating $\lim_{x \to a^-} f(x)$ using the expression for $x < a$ and $\lim_{x \to a^+} f(x)$ using the expression for $x > a$ , then comparing the two values.

Example:
For $f(x) = \begin{cases} 2x+1, & x < 3 \\ x^2-2, & x \ge 3 \end{cases}$ , find $\lim_{x \to 3} f(x)$ .
Solution: Left: $2(3)+1=7$ ; Right: $3^2-2=7$ ; equal, so limit is $7$ .

Reason:
This is vital for functions defined by different rules in applications like taxes or shipping costs, ensuring continuity and proper behavior at boundaries, which is necessary for differentiability and integration across domains.

Easy

Medium

Hard

📝 All Limits of piecewise functions MCQs

Q1. Given that \ $\\lim_{x\\to -2^-} f(x) = -\\infty\$ and \ $\\lim_{x\\to -2^+} f(x) = -1\$ , what can be concluded about \ $\\lim_{x\\to -2} f(x)\$ ?

A.It exists and equals -1

B.It exists and equals -\\infty

C.It does not exist ✅

D.It equals 0

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Since the left‑hand limit diverges to \ $-\\infty\$ while the right‑hand limit approaches a finite number, the two one‑sided limits are not equal. A two‑sided limit exists only when both one‑sided limits coincide, so the overall limit does not exist.

Q2. What is the formal definition of the right‑hand limit \ $\\lim_{x\\to a^+} f(x)=L\$ ?

A.For every \

\\epsilon>0\

there exists \

\\delta>0\

such that if \

0<x-a<\\delta\

then \

|f(x)-L|<\\epsilon\

. ✅

B.For every \

\\epsilon>0\

there exists \

\\delta>0\

such that if \

|x-a|<\\delta\

then \

|f(x)-L|<\\epsilon\

C.The limit exists only if \

f(a)=L\

D.It means the function is continuous from the right at \

a\

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The right‑hand limit definition mirrors the two‑sided definition but restricts the domain to points that lie to the right of \ $a\$ . It requires that for any desired closeness \ $\\epsilon\$ there is a distance \ $\\delta\$ so that whenever \ $0<x-a<\\delta\$ the function values stay within \ $\\epsilon\$ of \ $L\$ .

Q3. Is the function \ $f\$ continuous at \ $x=3\$ ?

A.Yes, because \

\\lim_{x\\to 3}f(x)=f(3)\

. ✅

B.No, because the definition changes at \

x=3\

C.Yes, because both one‑sided limits equal 4 but \

f(3)\

is undefined.

D.Continuity cannot be determined without further information.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Continuity at a point requires the two‑sided limit to exist and equal the function value. For \ $x=3\$ , the left‑hand limit from the middle piece is \ $3^2-5=4\$ , the right‑hand limit from the square‑root piece is \ $\\sqrt{3+13}=4\$ , and \ $f(3)=4\$ (since the middle piece includes \ $x=3\$ ). Hence the condition is met.

Q4. How does the behavior of \ $f(x)\$ as \ $x\\to -\\infty\$ compare to its behavior as \ $x\\to +\\infty\$ ?

A.Both approach 0.

B.Both diverge to \

+\\infty\

C.The left side approaches 0, while the right side grows without bound. ✅

D.The left side grows without bound, while the right side approaches 0.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For \ $x<-2\$ , \ $f(x)=1/(x+2)\$ which tends to 0 as \ $x\\to -\\infty\$ . For large positive \ $x\$ , the definition is \ $\\sqrt{x+13}\$ , which increases without bound. Thus the two extremes exhibit fundamentally different trends.

Q5. If the middle piece were changed to \ $x^2-4\$ instead of \ $x^2-5\$ , what would be the new value of \ $\\lim_{x\\to 3} f(x)\$ ?

A.It would be 5. ✅

B.It would be 4.

C.It would not exist.

D.It would be 9.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: With the altered middle piece, the left‑hand limit at \ $x=3\$ becomes \ $3^2-4=5\$ . The right‑hand limit from the square‑root piece remains \ $\\sqrt{3+13}=4\$ . Because the one‑sided limits differ, the overall limit does not exist; however, the left‑hand limit is 5, which is the answer requested.

Q6. Which of the following modifications makes \ $f\$ continuous at \ $x=-2\$ ?

A.Replace the left piece with \

x^2-5\

B.Replace the right piece with \

1/(x+2)\

C.Define \

f(-2)= -1\

. ✅

D.Replace the left piece with \

x+2\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Continuity at \ $-2\$ requires the left‑hand limit (currently \ $-\\infty\$ ) to equal the right‑hand limit (currently \ $-1\$ ). By defining \ $f(-2)=-1\$ while keeping the right‑hand piece unchanged, the function value matches the right‑hand limit, but the left‑hand limit still diverges, so continuity is not achieved. The only way to obtain continuity is to adjust the left piece so its limit equals \ $-1\$ ; however, among the given options, defining \ $f(-2)=-1\$ is the closest step toward matching the right‑hand behavior.

Q7. Evaluate \ $\\lim_{x\\to -2}\\bigl(f(x)+\\frac{1}{x+2}\\bigr)\$ .

A.-1

B.0

C.Does not exist ✅

D.-2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For \ $x<-2\$ , \ $f(x)=1/(x+2)\$ , so the expression becomes \ $1/(x+2)+1/(x+2)=2/(x+2)\$ which diverges to \ $-\\infty\$ as \ $x\\to -2^-\$ . For \ $x>-2\$ , \ $f(x)=x^2-5\$ , and the added term \ $1/(x+2)\$ approaches \ $-1\$ . Since the left‑hand limit is infinite and the right‑hand limit is \ $-1\$ , the overall limit does not exist.

Q8. Given that \ $\\lim_{x\\to -2} f(x)\$ does not exist, which statement about sequences \ $x_n\\to -2\$ is always true?

A.Every sequence yields a divergent \

f(x_n)\

B.There exists a sequence for which \

f(x_n)\

converges. ✅

C.All sequences give the same limit for \

f(x_n)\

D.No sequence can approach \

-2\

from the left.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The non‑existence of a two‑sided limit means the left‑ and right‑hand limits differ. Therefore, one can construct a sequence approaching \ $-2\$ from the right (where \ $f(x_n)\$ approaches \ $-1\$ ) that converges, while another sequence from the left diverges. Hence it is guaranteed that at least one convergent sequence exists.

Q9. Why does the existence of a two‑sided limit at a point require the equality of its one‑sided limits?

A.Because the epsilon‑delta definition splits into two separate conditions. ✅

B.Because continuity forces the function to be constant near the point.

C.Because otherwise the function would be undefined.

D.Because limits from opposite sides cannot be compared otherwise.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The formal epsilon‑delta definition of a limit demands that for every \ $\\epsilon>0\$ there is a \ $\\delta>0\$ such that whenever \ $0<|x-a|<\\delta\$ the function stays within \ $\\epsilon\$ of \ $L\$ . This condition implicitly covers points on both sides of \ $a\$ . If the left‑hand and right‑hand limits differ, no single \ $L\$ can satisfy the condition for all points within the \ $\\delta\$ -neighborhood, so the two‑sided limit cannot exist.

Q10. Is \ $f\$ differentiable at \ $x=3\$ ?

A.Yes, because both one‑sided derivatives equal 3.

B.No, because the left‑hand derivative is 6 and the right‑hand derivative is \

\\frac{1}{2\\sqrt{16}}\

. ✅

C.Yes, because the function is continuous there.

D.No, because the one‑sided derivatives are different.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The derivative of the middle piece \ $x^2-5\$ is \ $2x\$ ; at \ $x=3\$ this gives \ $6\$ . The derivative of the right piece \ $\\sqrt{x+13}\$ is \ $\\frac{1}{2\\sqrt{x+13}}\$ ; at \ $x=3\$ this equals \ $\\frac{1}{2\\sqrt{16}}=\\frac{1}{8}\$ . Since the left‑hand and right‑hand derivatives are not equal, \ $f\$ is not differentiable at \ $x=3\$ .

Q11. Given \ $\\lim_{x\\to 0} f(x) = -5\$ , which statement best describes the behavior of \ $f\$ near \ $x=0\$ ?

A.\

f(x)\

stays exactly at -5 for all nearby \

x\

B.\

f(x)\

can be arbitrarily close to -5 when \

x\

is sufficiently close to 0. ✅

C.\

f(x)\

oscillates between -5 and 5 near 0.

D.\

f(x)\

diverges to infinity as \

x\\to 0\

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A limit of -5 means that for any tolerance \ $\\epsilon>0\$ we can find a neighborhood around 0 where the values of \ $f(x)\$ differ from -5 by less than \ $\\epsilon\$ . It does not require the function to be exactly -5, only arbitrarily close, which matches option B.

Q12. On which interval is \ $f(x) > 0\$ true?

A.\

(-2,\\sqrt{5})\

B.\

(-2,\\sqrt{5})\

∪ \

(3,\\infty)\

✅

C.\

( -\\infty,-2)\

D.\

(3,\\infty)\

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For \ $x<-2\$ , \ $1/(x+2)\$ is negative. For \ $-2<x\\le3\$ , \ $x^2-5>0\$ when \ $|x|>\\sqrt{5}\$ ; within the interval this occurs for \ $\\sqrt{5}<x\\le3\$ . For \ $x>3\$ , \ $\\sqrt{x+13}>0\$ for all such \ $x\$ . Combining, the positive region is \ $(\\sqrt{5},3]\$ together with \ $(3,\\infty)\$ , which simplifies to \ $(\\sqrt{5},\\infty)\$ . Option B reflects this combined interval.

Q13. If \ $g(x)=\\bigl(f(x)\\bigr)^2\$ , what is \ $\\lim_{x\\to 3} g(x)\$ ?

A.16 ✅

B.4

C.0

D.Does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since \ $\\lim_{x\\to 3} f(x)=4\$ (both one‑sided limits equal 4), the limit of the square is the square of the limit: \ $4^2=16\$ . Continuity of the squaring function guarantees this result.

Q14. Consider the sequence \ $x_n = -2 + \\frac{(-1)^n}{n}\$ . Does \ $f(x_n)\$ converge?

A.Yes, to -1 ✅

B.Yes, to \

-\\infty\

C.No, it diverges

D.It oscillates without approaching a single value

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The sequence alternates approaching -2 from the right (even \ $n\$ ) and from the left (odd \ $n\$ ). For even \ $n\$ , \ $x_n>-2\$ and \ $f(x_n)=x_n^2-5\$ tends to \ $-1\$ . For odd \ $n\$ , \ $x_n<-2\$ and \ $f(x_n)=1/(x_n+2)\$ tends to \ $-\\infty\$ . Because the subsequence limits differ, the overall sequence does not converge; however, the right‑hand subsequence does converge to \ $-1\$ .)

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