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BITSAT 2022 Paper

Section: Mathematics

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Question : 117 of 130

Marks: +1, -0

The value of

\lim\limits_{x\to 0}\;\frac{(1+x)^{\frac{1}{x}}-e+\frac{1}{2}ex}{x^2}

$\;\frac{11}{24}e$
$-\;\frac{11}{24}e$
$\;\frac{e}{24}$
None of these

Solution: 👈: Video 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

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]

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

[\because O(x)

in terms containing

x]

lim

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

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