Question

What is: ?

`""_(xrarr0)^lim (e^x-1)/(log(1+2x))`

Answer 1

`Lim_(x->0)((e^x-1)/log(1+2x))=ln10/2`

Explanation:

.

`Lim_(x->0)((e^x-1)/log(1+2x))`

We use the L'Hopital's Rule and take the limit of the derivatives of the numerator and the denominator:

`d/dx(e^x-1)=e^x`

`d/dx(log(1+2x))=d/dx(ln(1+2x)/ln10)=1/ln10*d/dx(ln(1+2x))=1/ln10*2/(1+2x)=2/(ln10(1+2x)`

`Lim_(x->0)((e^x-1)/log(1+2x))=Lim_(x->0)(e^x/(2/(ln10(1+2x))))=Lim_(x->0)((e^x(1+2x)ln10)/2)=ln10/2`