To prove the limit is e you do the following
$$
L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n
$$
\begin{align}
\ln L
&= \lim_{n \to \infty} \ln \left(1 + \frac{1}{n} \right)^n \\
&= \lim_{n \to \infty} n \ln \left(1 + \frac{1}{n} \right) \\
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n},
\end{align}
which you can evaluate with L'hopital's rule (take derivative of top and bottom, since both
go towards 0):
\begin{align}
\ln L
&= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n} \\
&= \lim_{n \to \infty} \frac{\frac{1}{1 + \frac{1}{n}}\left(\frac{-1}{n^2}\right)} {\frac{-1}{n^2}}\\
&= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}\\
&= 1.
\end{align}
Since the natural log of your limit is $1$, the limit itself must be $e$.$$$$
I cant understand the second statement in which we take the log of both sides and switch the limit and ln . The argument for doing such a thing is that the limit exists and the function is continuous (log is continuous) but how do we know the limit exists .
So don’t we need to prove
$$
L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n
$$
Exits before showing the above proof (and how do we prove it exists) or am i missing something
Best Answer
The sequence $\left(1+\frac1n\right)^n$ is increasing and bounded above (by $4$). Therefore, it converges. The remaining problem is therefore to find its limit.