Use Taylor Series to show that $\displaystyle \lim_{x \longrightarrow 0} \frac{\ln(1+x)}{x}=1.$
Ok, this is very intuitive, we have $\ln(1+x) = \displaystyle \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}$,
and $$ \lim_{x\rightarrow 0}\frac{\ln(1+x)}{x}= \lim_{x \rightarrow 0} \frac{\displaystyle\sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}}{x}= \lim_{x \rightarrow 0} \displaystyle\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n-1}}{n}= \lim_{x \rightarrow 0} \hspace{0.2cm} 1-\frac{x}{2} + \frac{x^2}{3}-\cdots = 0.$$
Ok, but in the last step I automatically assumed that the limits are comutatives, it's correct? Why?