I saw this proof that $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e$:
Let $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = L$. Then
$\log\left(\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \right) = \log(L) \iff
\lim_{n \to \infty} \left(\log \left(1+\frac{1}{n}\right)^n \right) = \log(L) \iff
\lim_{n \to \infty} \left(n \cdot log\left(1+\frac{1}{n}\right)\right) = \log(L) \iff
\lim_{n \to \infty} \left(\frac{log\left(1+\frac{1}{n}\right)}{\frac{1}{n}} \right) = \log(L)$. This leads us to $\left(\frac{0}{0}\right)$, so we can use L'Hôpital's rule to get $\lim_{n \to \infty} \left( \frac{\frac{-\frac{1}{n^2}}{1+\frac{1}{n}}}{-\frac{1}{n^2}}\right) = \log(L) \iff
\lim_{n \to \infty} \left(\frac{1}{1+\frac{1}{n}}\right) = \log(L) \iff 1=\log(L)$, from which we would conclude that $L=e$.
I was thinking that this same proof would work if instead of using the natural logarithm, I use other base. If I used $\log_{10}$, for example, I would get that the limit is 10, which can't happen because of the uniqueness of the limit.
What am I missing?
Best Answer
Recall that $(\log_a x)'=\frac1x \log_a e$ and therefore using $\log_{10}$ we obtain
$$\left(1+\frac{1}{n}\right)^n =10^\frac{\log_{10}\left(1+\frac1n\right)}{\frac1n} \to 10^{\log_{10}e}=e$$
indeed by l'Hospital
$$\lim_{n\to \infty }\frac{\log_{10}\left(1+\frac1n\right)}{\frac1n}=\lim_{n\to \infty } \frac{-\frac{n}{n+1}\frac1{n^2}\log_{10} e}{-\frac1{n^2}}= \log_{10}e$$