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