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    JEE Main Session 2 Answer Key 2026 Link: Live Updates, Download PDF at jeemain.nta.nic.in

    Limit Using Expansion - Practice Questions & MCQ

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

    Quick Facts

    • 117 Questions around this concept.

    Solve by difficulty

    \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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    Books

    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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