BPSC 45th Exam Prelims 2003 Solved Paper

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Question : 116 of 150

Marks: +1, -0

{\text "lim"}↙{ x →0} { xe^x -log[1+x] }/{x^2}

is equal to

$1/ 2$
$3/ 2$
0
$5/ 2$

Solution:

{\Lt}↙{x → 0} \;{x e^{x}-\log (1+x)}/{x^{2}}

Using L' Hospital rule
if

{Lt}↙{x → a} \;{f(x)}/{g(x)}=\;{0}/{0}

or

{\Lt}↙{x → a} \;{f(x)}/{g(x)}=\;{± ∞}/{± ∞}

then

{\Lt}↙{x → a} \;{f(x)}/{g(x)}={\Lt}↙{x → a} \;{f^{'}(x)}/{g^{'}(x)}

So,

={\Lt}↙{x → 0} \;{x e^{x}+e^{x}-{1}/{1+x}}/{2 x}

Again, By L' Hospital Rule

={\Lt}↙{x → 0} \;{e^{x}+x e^{x}+e^{x}+{1}/{(1+x)^{2}}}/{2}

Putting

x=0

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

=\;{1+1+1}/{2}=\;{3}/{2}

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