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BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Limits

BITSAT 2022 · Mathematics · Limits

The value of _ x arrow 0 (1+x)^ (1)/(x) -e+(1)/(2) e x x^2 is

The value of

\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e+\frac{1}{2} e x}{x^2}

is

Options

A $\frac{11}{24} e$
B $-\frac{11}{24} e$
C $\frac{e}{24}$
DNone of these

Correct answer

A. $\frac{11}{24} e$

Step-by-step solution

Let

y=(1+x)^{1 / x}\Rightarrow \log y=\frac{1}{x} \log (1+x)=\frac{1}{x}\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right]=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots

\begin{aligned} y & =e^{\left(1-\frac{x}{2}+\frac{x^2}{3}+\ldots.\right)}=e e^{\left(\frac{-x}{2}+\frac{x^2}{3} \ldots..\right)} \\ & =e\left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)+\frac{1}{2 !}\left(-\frac{x}{2}+\frac{x^2}{3}-\ldots\right)^2+\ldots\right]\end{aligned}

y-e+\frac{1}{2} e x=e x^2\left[\frac{1}{3}+0(x)+\frac{1}{2}\left(-\frac{1}{2}+0(x)\right)^2+\ldots.\right][\because 0(x)

in terms containing

x]\lim \frac{y-e+\frac{1}{2} e x}{x^2}=e\left[\frac{1}{3}+\frac{1}{8}\right]=\frac{11}{24} e

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