Amity University Noida-B.Tech Admissions 2026
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
117 Questions around this concept.
is
JEE Main 2026: Rank Predictor | College Predictor
Comprehensive Guide: IIT's | NIT's | IIIT's | Foreign Universities in India
Don't Miss: India's Best B.Tech Counsellors in your city - Book Your Seat
Find
The value of so that the fanction
becomes continuous is equal to
The value of $ \lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e}{x} $ is
$
\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots .+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots . .+n^3\right)-\left(1^2+2^2+\ldots . .+n^2\right)}
$
is equal to:
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Last Date to Apply: 29th April | Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements
If $\lim_{x\to 0}\frac{\ln \left ( 1+x \right )+\alpha \sin x+\frac{x^{2}}{2}}{x^{3}}= \beta \left ( \in R \right )$ ,
then the value of $\alpha +\beta$ is
Define a sequence $\left(a_n\right)$ by $a_1=5, a_n=a_1 a_2 \ldots \ldots a_{n-1}+4$ for $n>1$. Then $\lim _{n \rightarrow \infty} \frac{\sqrt{a_n}}{a_{n-1}}$
Define a sequence $\left\{a_n\right\}_{n \geq 0}$ by $a_n=\sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1, a_0=\cos \theta \neq 1$. Then $\lim _{n \rightarrow \infty} 4^n\left(1-a_n\right)$ equals
Limit Using Expansion (Part 1)
Using the expansions is one of the methods to find the limits. The following expansion formulas which are also known as Taylor series, are very useful in evaluating various limits.
(i) $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \ldots$.
(ii) $a^x=1+x\left(\log _e a\right)+\frac{x^2}{2!}\left(\log _e a\right)^2+\frac{x^3}{3!}\left(\log _e a\right)^3+\ldots \ldots \ldots$
(iii) $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \ldots \ldots$.
(iv) $\log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots \ldots \ldots$.
(v) $(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots \ldots$.
"Stay in the loop. Receive exam news, study resources, and expert advice!"