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📝 Left hand and right hand limits (13 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 13 questions available

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What is Left hand and right hand limits?

Definition:
A left-hand limit, denoted $\lim_{x \to a^-} f(x)$ , examines the value $f(x)$ approaches as $x$ approaches $a$ from values less than $a$ (from the left). Conversely, a right-hand limit, $\lim_{x \to a^+} f(x)$ , considers values greater than $a$ (from the right). These one-sided limits help analyze discontinuities and piecewise behavior near $a$ .

Example:
For $f(x) = \begin{cases} x+1, & x < 2 \\ x^2, & x \ge 2 \end{cases}$ , find left and right limits at $2$ .
Solution: $\lim_{x \to 2^-} f(x) = 2+1=3$ , and $\lim_{x \to 2^+} f(x) = 2^2=4$ .

Reason:
These concepts are critical for identifying jumps or breaks in functions and for determining the existence of the two-sided limit, which requires both one-sided limits to be equal, ensuring smooth transitions in applications.

Easy

Medium

Hard

📝 All Left hand and right hand limits MCQs

Q1. What does the notation $\lim_{x \to a^{+}} f(x) = L$ represent?

A.The limit of f(x) as x approaches a from the right ✅

B.The limit of f(x) as x approaches a from the left

C.The two‑sided limit of f(x) as x approaches a

D.The derivative of f at x = a

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The notation indicates the right‑hand, or one‑sided, limit of the function f as x approaches a from values greater than a. It states that the function values can be made arbitrarily close to the number L by taking x sufficiently close to a on the right side. This is distinct from left‑hand or two‑sided limits.

Q2. If $\lim_{x \to 2^{-}} g(x) = 5$ and $\lim_{x \to 2^{+}} g(x) = 5$ , which of the following must be true about $\lim_{x \to 2} g(x)$ ?

A.It does not exist

B.It could be any number

C.It equals 5 ✅

D.It equals 0

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When both the left‑hand and right‑hand limits at a point exist and equal the same number, the two‑sided limit exists and equals that number. Therefore, the overall limit as x approaches 2 must be 5. The other choices contradict this fundamental property of limits.

Q3. Suppose $f(x) = \frac{x^2-4}{x-2}$ for $x \neq 2$ . Which statement correctly describes $\lim_{x \to 2^{-}} f(x)$ and why?

A.The limit does not exist because the function is undefined at x=2.

B.The limit equals 4 because the expression simplifies to

x+2

for

x \neq 2

. ✅

C.The limit equals 0 because numerator approaches 0.

D.The limit is infinite because denominator approaches 0.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression $\frac{x^{2}-4}{x-2}$ simplifies to $x+2$ for all x ≠ 2. As x approaches 2 from the left, $x+2$ approaches 4. Hence the left‑hand limit exists and equals 4. The other options misinterpret the behavior of the simplified form or ignore the removable discontinuity.

Q4. Given the piecewise function $h(x) = \begin{cases} 3x+1, & x<1 \\ 2x-4, & x\ge 1 \end{cases}$ , what is $\lim_{x \to 1^{-}} h(x)$ and what does it imply about continuity from the left?

A.The left-hand limit is 5, indicating left continuity.

B.The left-hand limit is 3, indicating a jump discontinuity.

C.The left-hand limit does not exist.

D.The left-hand limit is 4, so the function is continuous from the left at x=1. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The piecewise definition gives $h(x)=3x+1$ for x<1. Taking the limit as x approaches 1 from the left yields $3(1)+1=4$ . Since the function’s value from the left matches this limit, the function is continuous from the left at x=1. The other statements either give incorrect limit values or misstate continuity.

Q5. If a function satisfies $\lim_{x \to a^{-}} f(x) = L$ and $\lim_{x \to a^{+}} f(x) = L$ , which of the following statements is always false?

A.The function must be discontinuous at x = a. ✅

B.Both one‑sided limits exist and are equal.

C.The two‑sided limit

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

exists and equals L.

D.The behavior of f(x) near a can be described by a single value.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: If both one‑sided limits exist and are equal to L, the function cannot be discontinuous at a; it is at least continuous in the limit sense. Therefore, the claim that the function must be discontinuous is always false. The remaining statements correctly describe properties that hold when the one‑sided limits agree.

Q6. Consider $p(x) = \begin{cases} \sin\frac{1}{x}, & x>0 \\ 0, & x\le 0 \end{cases}$ . Which of the following best explains why $\lim_{x \to 0^{+}} p(x)$ does not exist?

A.Because the function is defined as 0 for x≤0.

B.Because as x→0⁺,

\sin\frac{1}{x}

oscillates between –1 and 1 without settling to a single value. ✅

C.Because the denominator approaches infinity.

D.Because the function has a removable discontinuity at 0.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: As x approaches 0 from the right, the argument of the sine function, $1/x$ , grows without bound, causing $\sin(1/x)$ to oscillate rapidly between –1 and 1. Because the values do not settle toward any single number, the right‑hand limit fails to exist. The other options either refer to the left side or mischaracterize the nature of the discontinuity.

Q7. Compare the one‑sided limits of $f(x)=|x|$ at $x=0$ . Which statement correctly reflects their relationship?

A.Both limits equal 0.

B.The right-hand limit is –1, the left-hand limit is 1.

C.Both limits are undefined.

D.The right-hand limit is 0 and the left-hand limit is 0, showing they are equal. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: For $f(x)=|x|$ , as x approaches 0 from either side, the absolute value yields non‑negative numbers that tend to 0. Both the left‑hand and right‑hand limits equal 0, showing they are identical. This demonstrates that the function is continuous at 0, contrary to any claim of differing limits.

Q8. Analyze the function $q(x)=\frac{1}{x}$ . Which of the following correctly describes the relationship between $\lim_{x \to 0^{-}} q(x)$ and $\lim_{x \to 0^{+}} q(x)$ ?

A.Both limits are infinite and have the same sign.

B.Both limits equal zero.

C.The left-hand limit is –∞ while the right-hand limit is +∞, indicating they are not equal. ✅

D.Both limits do not exist for the same reason.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The function $1/x$ diverges to negative infinity when x approaches 0 from the left, because the denominator is negative and its magnitude shrinks. From the right, the values grow to positive infinity. Since the signs differ, the one‑sided limits are not equal, and the overall limit does not exist.

Q9. For the piecewise function $r(x)=\begin{cases} x^2, & x\le 3 \\ 6x-9, & x>3 \end{cases}$ , evaluate $\lim_{x \to 3^{-}} r(x)$ and compare it to $\lim_{x \to 3^{+}} r(x)$ . Which answer is correct?

A.Left limit is 9, right limit is 9; they are equal. ✅

B.Left limit is 9, right limit is 12; they differ.

C.Left limit does not exist, right limit is 9.

D.Both limits are infinite.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For x ≤ 3, $r(x)=x^{2}$ gives a value of $3^{2}=9$ as x approaches 3 from the left. For x>3, $r(x)=6x-9$ gives $6(3)-9=9$ as x approaches 3 from the right. Both limits equal 9, so they are the same, indicating no jump at x=3.

Q10. Given $s(x)=\sqrt{x}$ defined for $x\ge 0$ , what can be inferred about $\lim_{x \to 0^{-}} s(x)$ and why?

A.The left-hand limit equals 0 because sqrt is defined for negative numbers.

B.The left-hand limit does not exist because the function is not defined for x<0. ✅

C.The left-hand limit equals 0 by continuity.

D.Both limits are undefined but the right-hand limit equals 0.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The square‑root function is defined only for non‑negative arguments. Hence there are no function values for x<0, making the left‑hand limit undefined. While the right‑hand limit as x→0⁺ exists and equals 0, the left side cannot be evaluated, so the left‑hand limit does not exist.

Q11. Consider the function $t(x)=\frac{x^2-9}{x-3}$ for $x\neq 3$ . Determine the limit $\lim_{x\to 3} t(x)$ using one‑sided limits and explain why both sides yield the same result.

A.Limits differ because numerator changes sign.

B.The limit does not exist due to a hole at x=3.

C.Each one‑sided limit approaches 6, but the overall limit is undefined.

D.Both one‑sided limits equal 6 because the expression simplifies to

x+3

, so the two‑sided limit exists and equals 6. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: For x ≠ 3, the expression simplifies to $x+3$ . Taking the limit as x approaches 3 from either side yields $3+3=6$ . Because both one‑sided limits equal 6, the two‑sided limit exists and is 6. The other options incorrectly claim divergence or undefined behavior.

Q12. Which principle explains why the existence of both one‑sided limits at a point guarantees the existence of the two‑sided limit?

A.Intermediate Value Theorem

B.Mean Value Theorem

C.If the left‑hand and right‑hand limits exist and are equal, then the two‑sided limit exists. ✅

D.Fundamental Theorem of Calculus

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The principle states that if both the left‑hand and right‑hand limits at a point exist and are equal to the same finite number, then the ordinary (two‑sided) limit exists and equals that number. This logical connection underlies many proofs involving continuity and limit evaluation.

Q13. Evaluate the statement: “If $\lim_{x\to a^{+}} f(x)$ exists and is finite, then $\lim_{x\to a} f(x)$ must also exist.” Which choice correctly assesses its validity?

A.False; the left‑hand limit may fail to exist or differ, preventing the two‑sided limit. ✅

B.True; the right‑hand limit alone determines the two‑sided limit.

C.True; existence of one‑sided limit guarantees continuity.

D.False; a function cannot have a right‑hand limit without a left‑hand limit.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The statement is false: a right‑hand limit alone does not guarantee the existence of the overall limit, because the left‑hand limit might be different or fail to exist. Both one‑sided limits must agree for the two‑sided limit to exist. Hence the correct assessment is that the claim is false.

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