Question

lim x tends to 0 e^ax - e^-ax/log(1+bx)
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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

Step-by-step explanation:

Given $\lim_{x \to 0} \frac{e^{ax} -e^{-ax} }{log(1+bx)}$

If we apply the limit, we get

= $\frac{e^{a*0} -e^{-a*0} }{log(1+b*0)}$

= $\frac{1 -1 }{log1}$

= $\frac{0}{0}$

As this is indeterminate form, we apply L Hospital's Rule

that is differentiating the numerator and denominator

=> $\lim_{x \to 0} \frac{e^{ax}*a -e^{-ax}*(-a) }{\frac{1}{ (1+bx)} *b}$                      (  $\frac{d}{dx}$ $e^{ax} = e^{ax}$ and $\frac{d}{dx}$ log x = $\frac{1}{x}$ )

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

Now applying the limits,

= $\frac{a(e^{0} +e^{0})(1+b*0) }{b}$

= $\frac{a(1+1)(1+0) }{b}$

= $\frac{a(2)(1) }{b}$

= $\frac{2a }{b}$

Therefore, $\lim_{x \to 0} \frac{e^{ax} -e^{-ax} }{log(1+bx)}$ = $\frac{2a }{b}$