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📝 Basic limits for constant and identity functions (12 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 12 questions available

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What is Basic limits for constant and identity functions?

Definition:
The limit of a constant function $f(x) = c$ as $x$ approaches any value $a$ is the constant itself, i.e., $\lim_{x \to a} c = c$ , because the output remains unchanged. The limit of the identity function $f(x) = x$ is the point of approach, $\lim_{x \to a} x = a$ , since the input and output coincide. These are fundamental building blocks for evaluating more complex limits.

Example:
Evaluate $\lim_{x \to 5} 7$ and $\lim_{x \to 5} x$ .
Solution: $\lim_{x \to 5} 7 = 7$ and $\lim_{x \to 5} x = 5$ .

Reason:
These basic limits serve as the foundation for applying limit laws, allowing us to break down complex functions into simpler components, which is essential for solving polynomial, rational, and other elementary function limits efficiently.

Easy

Medium

Hard

📝 All Basic limits for constant and identity functions MCQs

Q1. According to theorem 1.2.1 (a), what is $\displaystyle \lim_{x\to a} k$ where $k$ is a constant?

A.k ✅

B.a

C.0

D.does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Theorem 1.2.1(a) states that the limit of a constant function equals the constant itself because the function’s value does not change as $x$ approaches $a$ . Therefore the limit exists for any real $a$ and is exactly $k$ , independent of the approaching point.

Q2. Suppose $\displaystyle \lim_{x\to a} f(x)=5$ and $\displaystyle \lim_{x\to a} g(x)=-3$ . What is $\displaystyle \lim_{x\to a} [f(x)+g(x)]$ ?

A.-2

B.8

C.2 ✅

D.undefined

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: By the sum law for limits, the limit of a sum equals the sum of the limits, provided each individual limit exists. Adding the given limits gives $5+(-3)=2$ . Hence the limit of the combined expression is $2$ , which appears as option C.

Q3. If $\displaystyle \lim_{x\to a} f(x)=4$ and $\displaystyle \lim_{x\to a} g(x)=\frac12$ , find $\displaystyle \lim_{x\to a} f(x)g(x)$ .

A.4

B.8

C.1

D.2 ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The product law for limits says the limit of a product equals the product of the limits when both limits exist. Multiplying the given limits yields $4 \times \frac12 = 2$ . Therefore the limit of the product is $2$ , which is listed as option D.

Q4. Which statement correctly describes $\displaystyle \lim_{x\to 0^-}\frac{1}{x}$ and $\displaystyle \lim_{x\to 0^+}\frac{1}{x}$ ?

A.both equal 0

B.left limit = -∞ and right limit = +∞ ✅

C.both equal ∞

D.left limit = +∞ and right limit = -∞

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: As $x$ approaches zero from the left, the denominator is negative and its magnitude becomes very small, making the quotient tend toward $-\infty$ . Approaching from the right gives a positive small denominator, so the quotient grows without bound toward $+\infty$ . This matches option B.

Q5. Given that $\displaystyle \lim_{x\to a} f(x)=L$ exists and $\displaystyle \lim_{x\to a} \frac{1}{f(x)}$ also exists, which statement must be true?

A.L \neq 0 ✅

B.L = 0

C.f(x) is bounded near a

D.none of the above

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For the reciprocal limit to exist, the denominator $f(x)$ cannot approach zero; otherwise the reciprocal would blow up or be undefined. Hence the limit $L$ must be a non‑zero real number. This necessity is captured by option A.

Q6. Find $\displaystyle \lim_{x\to 9} \sqrt{x}$ .

A.9

B.3 ✅

\sqrt{9}

D.does not exist

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The square‑root function is continuous for all non‑negative arguments. Therefore the limit as $x$ approaches 9 equals the function value at 9, which is $\sqrt{9}=3$ . Option B correctly gives this value.

Q7. If $\displaystyle \lim_{x\to a} h(x)=7$ , what is $\displaystyle \lim_{x\to a} 3h(x)$ ?

A.21 ✅

B.7

C.3

D.undefined

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The constant‑multiple rule for limits states that multiplying a function by a constant multiplies its limit by the same constant. Hence $\displaystyle \lim_{x\to a} 3h(x)=3\cdot7=21$ . This result appears as option A.

Q8. Using theorem 1.2.2(e), which of the following limits is valid? $\displaystyle \lim_{x\to 4} \sqrt{x-1}$ .

A.3

B.2

\sqrt{3}

✅

D.does not exist

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The even‑root law requires the radicand to be positive near the limit point. Here $x-1$ approaches $3>0$ , so the limit equals the square root of the limit of the radicand: $\sqrt{4-1}=\sqrt{3}$ . Option C expresses this value.

Q9. If $\displaystyle \lim_{x\to a} p(x)=2$ and $\displaystyle \lim_{x\to a} q(x)=5$ , find $\displaystyle \lim_{x\to a} [p(x)q(x)+p(x)]$ .

A.14

B.7

C.10

D.12 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: First compute the product limit: $2\cdot5=10$ . Then add the limit of $p(x)$ , which is $2$ . So the overall limit is $10+2=12$ . This result matches option D.

Q10. Consider $g(x)=\frac{1}{k}$ where $k\neq 0$ is constant. What is $\displaystyle \lim_{x\to a} g(x)$ ?

A.0

\frac{1}{k}

✅

C.k

D.undefined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since $g(x)$ does not depend on $x$ at all, its value is the constant $\frac{1}{k}$ for every $x$ . Consequently the limit as $x$ approaches any point $a$ is the same constant, $\frac{1}{k}$ , which is option B.

Q11. Let $f(x)=\begin{cases}x^{2}& x<2\\4& x\ge 2\end{cases}$ . What is $\displaystyle \lim_{x\to 2} f(x)$ ?

A.2

B.does not exist

C.0

D.4 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Approaching 2 from the left, $f(x)=x^{2}$ gives a limit of $2^{2}=4$ . Approaching from the right, the function is constantly 4, so the right‑hand limit is also 4. Because both one‑sided limits agree, the two‑sided limit exists and equals 4, option D.

Q12. Given $\displaystyle \lim_{x\to 0} f(x)=0$ with $f(x)\ge 0$ near 0, what is $\displaystyle \lim_{x\to 0} \sqrt{f(x)}$ ?

A.does not exist

B.0 ✅

C.1

D.undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the limit of a non‑negative function is zero, the even‑root law applies because the radicand stays non‑negative near the limit point. The limit of the square root equals the square root of the limit: $\sqrt{0}=0$ . Hence the limit is 0, option B.

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