Proving that $\left(1+\lambda/n\right)^n\to e^\lambda$ I wish to show that $\lim_{n\to\infty}\left(1+\frac{\lambda}{n}\right)^n=e^\lambda$ as part of the proof of the Poisson limit theorem. My first avenue of attack was to take the log of both sides and then use L'Hopital's rule, but the result was of the form $0\cdot \infty$, so that didn't work.
I also noticed that the proof would follow from the proof that $\left(1+\frac{1}{n}\right)^\lambda = \left(1+\frac{\lambda}{n}\right)$ for large $n$, but I wasn't able to prove this either.
Any hints?
 A: If $$y=\left(1+\frac \lambda n\right)^n$$
so, $$\ln y=n\ln\left(1+\frac \lambda n\right)=\frac{\ln\left(1+\frac \lambda n\right)}{\frac1n}$$
so, $$\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{\ln\left(1+\frac \lambda n\right)}{\frac1n}\text{ which is in the form } \frac00 $$ 
Applying L'Hospital's rule, $$\lim_{n\to\infty}\ln y=\lim_{n\to\infty}\frac{-\frac \lambda {n^2}}{-\frac1{n^2}\left(1+\frac \lambda n\right)}=\lambda$$
So, $$\lim_{n\to\infty}\ln y=\lambda\iff \lim_{n\to\infty}=e^\lambda$$
A: The simple and straightforward way is to use substitution $n=\lambda t$. Your limit becomes:
$$\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^{\lambda t}=\left(\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t\right)^\lambda=e^\lambda$$
A: The answer very much depends on what you know, but, if this includes the fact that log is a primitive of $x\mapsto1/x$ and the inverse of $x\mapsto\mathrm e^x$, you may start from
$$
1-x\leqslant\frac1{1+x}\leqslant 1.
$$
Integrating this from $0$ to some $x\geqslant0$ yields
$$
x-\tfrac12x^2\leqslant\log(1+x)\leqslant x,
$$
which implies
$$
\mathrm e^{x-x^2/2}\leqslant1+x\leqslant\mathrm e^x,
$$
and, for every nonnegative $n$,
$$
\mathrm e^{nx-nx^2/2}\leqslant(1+x)^n\leqslant\mathrm e^{nx}.
$$
Using this for $x=\lambda/n$ yields the nonasymptotic bounds
$$
\mathrm e^{\lambda-\lambda^2/(2n)}\leqslant(1+\lambda/n)^n\leqslant\mathrm e^{\lambda}.
$$
Since $\lambda^2/(2n)\to0$, you are done.
A: $$e = \lim_{n \to \infty} \left(1+\frac 1 n\right)^n$$
$$e^\lambda =
\lim_{n \to \infty}  \left(1+\frac 1 n\right)^{n \lambda} =\\
\lim_{n \to \infty} \left( \left(1+\frac 1 n\right)^\lambda\right)^{n}=\\
\lim_{n \to \infty} \left(1+\frac{\lambda}{n}+O\left(\frac{1}{n^2}\right)\right)^{n}
$$
A: We know, $$e^y=\sum_{0\le r<\infty}\frac{y^r}{r!}$$
The $r$th term $T_r$ of $\left(1+\frac\lambda n\right)^n$ is 
$\frac {n(n-1)(n-2)\cdots\{n-(r-1)\}}{1\cdot2\cdot3\cdots (r-1)r}\left(\frac\lambda n\right)^r$ for $r\ge 1, T_0=1$ 
So, $\lim_{n\to\infty}T_r$ of $\left(1+\frac\lambda n\right)^n$ is 
$$\lim_{n\to\infty}\frac {n(n-1)(n-2)\cdots(n-r+1)}{1\cdot2\cdot3\cdots (r-1)r}\left(\frac\lambda n\right)^r$$
$$=\frac{\lambda^r}{r!}\lim_{n\to\infty}\prod_{0\le s\le r-1}\left(1-\frac sn\right)=\frac{\lambda^r}{r!}$$
So, $$\lim_{n\to\infty}\left(1+\frac\lambda n\right)^n=1+\sum_{1\le r<\infty}\frac{\lambda^r}{r!}=\sum_{0\le r<\infty}\frac{\lambda^r}{r!}=e^\lambda$$
