Question

( lim _{x rightarrow 0} frac{a e^{x}-b cos x+c e^{-x}}{x sin x}=2 )
at ( x=0 ) As linot exists and denuonimator - o suilth no numerator must tond to 0
At ( x=0 )
[
begin{array}{l}
a e^{0}-b cos (0)+c e^{-0}=0
Rightarrow a-b+c=0
end{array}
]
( lim _{x rightarrow 0} frac{a e^{x}-b cos x+c e^{-x}}{x^{2}(sin x)} frac{x}{x} )
( operatorname{asin} frac{sin x}{x} rightarrow 1 ) at ( x=0 ) it can me eliminatia
( =lim _{x rightarrow 0} frac{a e^{x}-b cos x+c e^{-x}}{x^{2}} )
Applying L Hopital's rule
( =lim _{x rightarrow 0} frac{a e^{x}+b sin x-c e^{-x}}{2 x} )
As demominator ( rightarrow 0 ) at ( x=0 )
[
9 e^{0}+b
]
( 1-e^{e^{0}}=0 )
Using (i) and (11) ( frac{a e^{x}+2 a operatorname{sen} x-a e^{-x}}{x}=2 )
[
=
]
( =lambdaleft(frac{e^{x}-1}{x}+frac{2 a sin x+aleft(frac{-x}{1}-1right)}{x}=2right. )
( 4 a=2 )
( Rightarrow quad 4 a=2 quad b=1 quad c=1 / 2 )

# It can also be solved by L'Hospital rule Find the value of a, b and c so that lim ae" - b cos x + cet -=2 xsin x Ex.8 10

Solution