VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
123 Questions around this concept.
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JEE Main Session 2 Memory Based Questions: April 2- Shift 1
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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:
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
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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) $(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots \ldots$.
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