Prove the limit of natural logarithm without differentiation or Taylor Series

[wolly]

Can someone help me to prove that this limit is equal to 1 without using differentiation or Taylor Series?(like in the limit of sinx/x when x tends to 0)What are the steps?

 

[HallsofIvy]

\(\displaystyle \lim_{x\to 0}\frac{ln(x+ 1)}{x}= \lim_{x\to 0}ln((x+ 1)^{\frac{1}{x}})= ln(\lim_{x\to 0} (x+1)^{\frac{1}{x}})= ln(e)= 1\)........... edited

 

[pka]

View attachment 12079
Can someone help me to prove that this limit is equal to 1 without using differentiation or Taylor Series?(like in the limit of sinx/x when x tends to 0)What are the steps?

\(\displaystyle \begin{align*}\lim_{x\to 0}\frac{ln(x+ 1)}{x}&= \lim_{x\to 0}ln((x+ 1)^{\frac{1}{x}}) \\&= \lim_{u\to \infty}ln((1+\frac{1}{u})^{u})\\&=\ln(e) \\&= 1\end{align*}\)

 

[HallsofIvy]

Thanks! I for some reason switched x multiplying rather than dividing!