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Limit Using Expansion - Practice Questions & MCQ

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

Quick Facts

  • 109 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 integer \mathrm{n} for which \mathrm{\lim _{x \rightarrow 0}\left\{\frac{(\cos x-1)\left(\cos x-e^2\right)}{x^n}\right\}} is a finite, non-zero real number, is


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

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\mathrm{f(x)} is the integral of \mathrm{\frac{2 \sin x-\sin 2 x}{x^3}, x \neq 0, find \lim _{x \rightarrow 0} f^{\prime}(x)}.

\mathrm{\lim _{n \rightarrow \infty} \cos \left[\pi \sqrt{\left(n^2+n\right)}\right] \text { when } n \text { is an integer, is equal to: } }

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The integer \mathrm{ n} for which  \mathrm{\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^x\right)}{x^n}} is a finite non-zero number is:

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 is also known as Taylor series, are very useful in evaluating various limits.

 \\\text { (i) } \quad e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots\\\\\text { (ii) } \quad 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\\\\ \text { (iii) } \quad \log (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots \ldots \ldots\\\\\text { (iv) } \quad \log (1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\ldots \ldots \ldots\\\\\text{ (v) } \quad(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)

\\\text { (vi) } \quad \sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots \ldots .\\\\\text { (vii) } \quad \cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\ldots \ldots .\\\\\text { (viii) } \quad \tan x=x+\frac{x^{3}}{3}+\frac{2}{15} x^{5}+\ldots \ldots .\\\\\text { (ix) } \quad \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\\\\\text { (x) } \quad \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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