limit

Question

solve following

$\lim_{x->0} e^{ax}- e^{-ax}/ log(1+bx)$

Comments

$\frac{2a}{b}$ using L-Hospital.

Answer 1

$\lim_{x\rightarrow 0}\frac{e^{ax}-e^{-ax}}{log(1+bx)}$ if we apply limit it become $\frac{0}{0}$ form

so we can apply l'hopital's rule differentiating numerator and denominator w.r.t x

$\lim_{x\rightarrow 0}\frac{ae^{ax}-(-ae^{-ax})}{\frac{b}{1+bx}}$

$\lim_{x\rightarrow 0}\frac{a(e^{ax}+e^{-ax})*(1+bx)}{b}$
applying limits we get

$\lim_{x\rightarrow 0}\frac{a(e^{ax}+e^{-ax})*(1+bx)}{b}$=$\frac{2a}{b}$

Comments

@Tesla!

it will be $\frac{2a}{b}$

when you are diffrentiating log(1+bx) w.r.t x, you will get $\frac{b}{1+bx}$

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Thanks @joshi_nitish

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thanks , i was doing mistakes in calculating