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Limit Using Expansion - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 124 Questions around this concept.

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\mathrm{ \lim _{x \rightarrow \infty}\left(1+\frac{3}{x}+\frac{5}{x^2}\right)^x \text { equals }}

\mathrm{\lim _{n \rightarrow \infty} \prod_{r=2}^n\left(\frac{r^3+1}{r^3-1}\right)} is
 

\mathrm{\lim _{x \rightarrow 0} \frac{3 e^x-x^3-3 x-3}{\tan ^2 x}}

Find  \mathrm{\lim _{x \rightarrow 0} \frac{(1+x)^{1 / m}-(1-x)^{1 / m}}{(1+x)^{1 / n}-(1-x)^{1 / n}}}

The value of \mathrm{f(0)}  so that the fanction \mathrm{f(x)= \frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}} 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:

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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 $\lim _{n \rightarrow \infty} \frac{\sqrt{a_n}}{a_{n-1}}$

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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

Concepts Covered - 2

Limit Using Expansion (Part 1)

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) $(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\frac{n(n-1)(n-2)}{3!} x^3+\ldots \ldots \ldots$.

Limit Using Expansion (Part 2)
Limit Using Expansion (Part 2) (vi) $\quad \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots \ldots \ldots$ (vii) $\quad \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots \ldots$. (viii) $\quad \tan x=x+\frac{x^3}{3}+\frac{2}{15} x^5+\ldots \ldots$. (ix) $\sin ^{-1} x=x+\frac{1^2}{3!} \cdot x^3+\frac{1^2 \cdot 3^2}{5!} \cdot x^5+\frac{1^2 \cdot 3^2 \cdot 5^2}{7!} \cdot x^7 \ldots \ldots \ldots$ (x) $\tan ^{-1} x=x-\frac{x^3}{3}+\frac{x^5}{5}+\ldots \ldots \ldots$.

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Limit Using Expansion (Part 1)
Limit Using Expansion (Part 2)

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Reference Books

Limit Using Expansion (Part 1)

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.15

Line : 22

Limit Using Expansion (Part 2)

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.15

Line : 30

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