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BPSC 45th Prelims 2003 Paper

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Question : 116 of 150

Marks: +1, -0

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

is equal to

$\frac{1}{2}$
$\frac{3}{2}$
0
$\frac{5}{2}$

Solution:

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

Using L' Hospital rule

if

\lim\limits_{x\to a} \frac{f(x)}{g(x)} = \frac{0}{0}

or

\lim\limits_{x\to a} \frac{f(x)}{g(x)} = \frac{\pm \infty}{\pm \infty}

then

\lim\limits_{x\to a} \frac{f(x)}{g(x)} = \lim\limits_{x\to a} \frac{f'(x)}{g'(x)}

So,

= \lim\limits_{x\to 0} \frac{x e^{x} + e^{x} - \frac{1}{1+x}}{2x}

Again, By L' Hospital Rule

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

Putting

x = 0

= \frac{e^{0} + 0 + e^{0} + \frac{1}{(1+0)^{2}}}{2}

= \frac{1+1+1}{2} = \frac{3}{2}

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