Limits, Continuity and Differentiability · Mathematics · BITSAT

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MCQ (Single Correct Answer)

1

The value of $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ is

BITSAT 2025
2

Consider the function $g(x)$ defined as

$$ g(x)=\left\{\begin{array}{cc} \frac{x^2-4}{x^2-2|x-2|-4}, & x \neq 2 \\ \frac{3}{4}, & x=2 \end{array}\right. $$

Which of the following statements is true about the continuity of $g(x)$ ?

BITSAT 2025
3

If $\mathop {\lim }\limits_{x \to \infty }\left\{\frac{x^2-1}{x+1}-a x-b\right\}=2$. The value of $a$ is

BITSAT 2025
4

Let $ f $ be the function defined by

$ f(x)=\left\{\begin{array}{cc} \frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\ \frac{1}{2}, & x=1 \end{array}\right. $

BITSAT 2024
5

The value of $$\lim _\limits{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}-\cos \frac{x^2}{4}+\cos \frac{x^2}{2} \cos \frac{x^2}{4}\right)$$ is

BITSAT 2023
6

$$ \text { The value of } \lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}} \text { equals to } $$

BITSAT 2023
7

The value of $$\mathop {\lim }\limits_{x \to 0} {{1-\cos (1 - \cos x)} \over {{x^4}}}$$ is

BITSAT 2022
8

The value of $$\mathop {\lim }\limits_{x \to 0} {{{{(1 + x)}^{{1 \over x}}} - e + {1 \over 2}ex} \over {{x^2}}}$$ is

BITSAT 2022
9

The value of $$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - {{\cos x}^2}} } \over {1 - \cos x}}$$ is

BITSAT 2021
10

If $$f(x) = \left\{ {\matrix{ {a{x^2} + 1,} & {x \le 1} \cr {{x^2} + ax + b,} & {x > 1} \cr } } \right.$$ is differentiable at x = 1, then

BITSAT 2021
11

The value of $$\mathop {\lim }\limits_{x \to \infty } {1 \over n}\left\{ {{1 \over {n + 1}} + {2 \over {n + 2}} + .... + {{3n} \over {4n}}} \right\}$$ is

BITSAT 2020