I need to solve the following limit without

Rachael Velasquez

Rachael Velasquez

Answered question

2022-04-06

I need to solve the following limit without using L'Hopital's rule:
limx0(1+sin(x))1x
The thing is that I can not figure out what to do. One of my ideas was to apply this rule: ax=eln(ax)=exln(a) , getting this:
limx0e1xln(1+sin(x))
I already know that the answer is e, so the exponent is definitely 1. However, I tried everything I could but have no idea how to solve limx0(1xln(1+sin(x))) which needs to be 1.
I would really appreciate your help, and if you find a totally different way to solve the limit without using L'Hospital's rule it will be good as well.

Answer & Explanation

Dallelopeelvep2yc

Dallelopeelvep2yc

Beginner2022-04-07Added 15 answers

Consider the function f(π2,π2)(1,1) defined by
f(x)=ln(1+sinx)
We have that f(0)=0, f is continuously differentiable, and f(x)=cosx1+sinx by the chain rule; so that
limx0ln(1+sinx)x=limx0f(x)f(0)x=f(0)=cos01+sin0=1
so that by continuity of exp
limx0e1xln(1+sinx)=elimx01xln(1+sinx)=e1=e

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