Mumbai University > FE > Semester 1 > Applied Mathematics 1

Marks : 08

Years : DEC 2015

Let $L=\lim\limits_{x⟶0} \dfrac {ae^x-be^{-x}+cx}{x-\sin x} =4…….(1) $

Since the limit L has a definite value it exists.

But as seen denominator will be zero at x=0 so numerator must also be zero at x=0

Putting x=0 in numerator

$∴\text{ we get } ae^0-be^0+c(0)=0 \\ ∴a-b=0 \\ ∴a=b……………..(2)$

Since Given limit is 0/0 type According to L Hospitals rule

$L=\lim\limits_{x⟶0} \dfrac {ae^x-be^{-x}(-1)+c(1)}{1-\cos x} \\ \lim\limits_{x⟶0} \dfrac {ae^x-ae^{-x}+c}{1-\cos x}…… (3)$

Since again denominator is 0 at x=0

∴Numerator must also be zero

$∴ae^x+ae^{-x}+c=0 \\ ∴a+a+c=0 \\ ∴2a=-c ……………….. (4)$

Again applying L Hospitals rule,

$\lim\limits_{x⟶0} \dfrac {ae^x-ae^{-x}(-1)+0}{+\sin x} \\ \therefore \lim\limits_{x\to0} \dfrac {ae^x-ae^{-x}}{\sin x} $

Again applying L Hospitals rule,

$\lim\limits_{x\to0} \dfrac {ae^x+ae^{-x}}{\cos x}=0 \\ ∴2a = 4 \\ ∴a=2 \\ B=2\space \space and\space\space c=-4.$