$d_n=(1+(2/n))^n$ converge or diverge and find the limit? I know the answer is $e^2$ and I'd like to use L'Hopital's rule because this is an indeterminate form. Can someone explain how to get there?
$$d_n=\left(1+\frac{2}{n}\right)^n$$
 A: Using the well known limit
$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$
we can get
$$\begin{align}
\lim_{n\to\infty}\left(1+\frac2n\right)^n & =\lim_{n\to\infty}\left(1+\frac1{n/2}\right)^n\\
& = \lim_{m\to\infty}\left(1+\frac1m\right)^{2m}\quad m=n/2\\
& = \lim_{m\to\infty}\left[\left(1+\frac1m\right)^m\right]^2\\
& = e^2
\end{align}$$
A: If you know that the sequence converges, then note that since the subsequence $d_{2n} = (1 + 1/n)^{2n} = [(1 + 1/n)^n]^2$ converges to $e^2$, so does $d_n$.
A: A term of the form $f(n)^{g(n)}$ can usually be converted to a L'Hopital's rule form by taking the log of both sides.
$$\ln d_n = \ln \left(\left(1+\frac 2n\right)^n\right) = n \ln \left(1+\frac 2n\right) = \frac{\ln\left(1+\frac 2n\right)}{\frac 1n}$$
From here you can probably show that
$$\lim_{n\rightarrow\infty} \ln d_n = 2.$$
Then, since $\ln$ is continuous,
$$\lim_{n\rightarrow\infty} \ln d_n = \ln \lim_{n\rightarrow\infty} d_n = 2,$$
and you can solve to get
$$\lim_{n\rightarrow\infty} d_n = e^2.$$
A: Apply $\ln$ to get
$$n \cdot \ln (1+2/n) =2\frac{\ln (1+2/n) -\ln 1}{2/n} \to 2\ln'(1) = 2.$$
No need for L'Hopital, just the definition of the derivative. Exponentiating back gives $e^2$ for the limit.
