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  • The value of  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e+\frac{e x}{2}}{x^2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Crightar

The value of  \mathrm{\lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e+\frac{e x}{2}}{x^2}} is

Option: 1

\frac{11 e}{24}


Option: 2

-\frac{11 e}{24}


Option: 3

\frac{e}{24}


Option: 4

none of these


Answers (1)

best_answer

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

\mathrm{(1+x)^{1 / x} }can be expanded as \mathrm{y=(1+x)^{1 / x} }

\mathrm{\log y=\frac{1}{x} \log (1+x)=\frac{1}{x}\left[x-\frac{x^2}{2}+\frac{x^3}{3}\right] }  [Higher power is neglected as \mathrm{\rightarrow 0 } ]

\mathrm{\Rightarrow \quad y=e^{1-\frac{x}{2}+\frac{x^2}{3}}=e\left[e^{-\frac{x}{2}+\frac{x^2}{3}}\right] }
Now from the expansion of exponential function
\mathrm{y=e\left[1+\left(\frac{-x}{2}+\frac{x^2}{3}\right)+\frac{1}{2 !}\left[\frac{x^2}{4}+\frac{x^4}{9}-\frac{2 x^3}{6}\right]+\ldots \ldots\right] }

Thus,
\mathrm{y=e-\frac{e x}{2}+e x^2\left[\frac{1}{3}+\frac{1}{8}\right]+\text { higher powers }}

\mathrm{Therefore, \lim _{x \rightarrow 0} \frac{(1+x)^{1 / x}-e+\frac{e x}{2}}{x^2}=\lim _{x \rightarrow 0} \frac{e-\frac{e x}{2}+x^2\left(\frac{11}{24}\right) e-e+\frac{e x}{2}}{x^2}=\frac{11 e}{24}}

Posted by

manish painkra

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