Difficulty distribution
How the classified questions are distributed by difficulty.
Your cart is empty.
Practice Limits Continuity And Differentiability - Calculus - Mathematics previous year questions organised from real papers, with year-wise coverage and clear topic navigation.
Every graph below is calculated only from this selection.
Year-wise coverage for Limits Continuity And Differentiability. Each bar uses a separate theme-derived color.
How the classified questions are distributed by difficulty.
MCQ, numerical, multiple-select and other formats found in these papers.
Top subjects by unique question coverage.
Top topics across the included previous year papers.
Top subtopics inside this exact selection.
Question coverage for the most populated papers. Every active PYP paper remains listed below.
Newest papers appear first. Sort by year, question coverage or name.
| Paper | Year / session | Questions in this view | Open |
|---|---|---|---|
| BITSAT 2025 | 2025 | 3 | View paper |
| BITSAT 2024 | 2024 | 1 | View paper |
| BITSAT 2023 | 2023 | 2 | View paper |
| BITSAT 2022 | 2022 | 2 | View paper |
| BITSAT 2021 | 2021 | 2 | View paper |
| BITSAT 2020 | 2020 | 1 | View paper |
Practice every matching question in batches of 20, with every available option.
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
The value of \(\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - {{\cos x}^2}} } \over {1 - \cos x}}\) is
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
The value of \(\mathop {\lim }\limits_{x \to 0} {{1-\cos (1 - \cos x)} \over {{x^4}}}\) is
The value of \(\mathop {\lim }\limits_{x \to 0} {{{{(1 + x)}^{{1 \over x}}} - e + {1 \over 2}ex} \over {{x^2}}}\) is
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
$$\text { The value of } \lim _\limits{x \rightarrow 0} \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}} \text { equals to }$$
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. $
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 $g(x)$ ?
If $\mathop {\lim }\limits_{x \to \infty }\left\{\frac{x^2-1}{x+1}-a x-b\right\}=2$. The value of $a$ is
The value of $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ is