On the justification of the result of a limit utilizing little-o. (quick question) Why is it that 
$$\lim_{n \rightarrow +\infty} \left( 1 + \frac{\lambda}{n}+ o \left( \frac{1}{n} \right) \right)^n = e^{\lambda}$$
I once remembered the justification for "ignoring" the $o(1/n)$ but can't recall it again.
 A: I suppose you know that $(1+(\lambda /n)) ^{n} \to e^{\lambda} $ and then you want to know why we can ignore $o(1/n)$. The simplest approach in my opinion is the one I saw on MSE itself based on the following lemma

Lemma: If $a_{n} $ is a sequence of real or complex terms such that $n(a_{n} - 1)\to 0$ ie $a_{n} =1+o(1/n)$ then $a_{n} ^{n} \to 1$.

Take $$a_{n} =\dfrac{1+\dfrac {\lambda} {n} +o(1/n)}{1+\dfrac{\lambda} {n}} $$ and you are done. 
A: The idea is that $o\!\left(\frac{1}{n}\right)$ is "just small enough enough" to counteract the power $n$.
Let's do the standard thing here, and write it with the exponential form:
$$
\left( 1 + \frac{\lambda}{n}+ o\!\left( \frac{1}{n} \right) \right)^n
= \exp \left( n\ln\left( 1 + \frac{\lambda}{n}+ o\!\left( \frac{1}{n} \right) \right)\right)
$$
Using the usual Taylor expansions of $\ln(1+u)=u+o(u)$ around $0$, we get
$$
\left( 1 + \frac{\lambda}{n}+ o\!\left( \frac{1}{n} \right) \right)^n
= \exp \left( n\left( \frac{\lambda}{n}+ o\!\left( \frac{1}{n} \right) \right)\right)
= \exp \left( \lambda + o\!\left( 1 \right) \right)
= e^{\lambda}e^{o(1)}
$$
and since $e^{o(1)}\xrightarrow[n\to\infty]{}e^0 = 1$, we get the result.
A: Use the definition:
$$f(n)\in o(g(n))\iff\forall \epsilon>0,\exists N\in\Bbb N: |f(n)|<\epsilon\cdot |g(n)|,\quad \forall n\ge N$$
Hence for each $\epsilon >0$ exists $N\in\Bbb N$ such that
$$\left( 1 + \frac{\lambda}{n}+ f(n) \right)^n\le\left( 1 + \frac{\lambda}{n}+ |f(n)| \right)^n<\left( 1 + \frac{\lambda+\epsilon}{n} \right)^n,\quad\forall n\ge N\tag{1}$$
and because for $\epsilon\in(0,1)$ we have that $-1<-\epsilon/n<-|f(n)|\le 0$ then
$$\left( 1 + \frac{\lambda-\epsilon}{n} \right)^n<\left( 1 + \frac{\lambda}{n}- |f(n)| \right)^n\le \left( 1 + \frac{\lambda}{n}+ f(n) \right)^n,\quad\forall n\ge N,\epsilon\in(0,1)\tag{2}$$
Then putting together $(1)$ and $(2)$ and taking the limit when $n\to\infty$ for any fixed $\epsilon\in(0,1)$ we have that
$$e^{\lambda-\epsilon}\le\lim_{n\to\infty}\left( 1 + \frac{\lambda}{n}+ f(n) \right)^n\le e^{\lambda+\epsilon}$$
Because the above holds for every $\epsilon\in(0,1)$ we can take limits again for $\epsilon\to 0^+$ to see that
$$e^\lambda=\lim_{\epsilon\to 0^+}e^{\lambda-\epsilon}\le\lim_{n\to\infty}\left( 1 + \frac{\lambda}{n}+f(n) \right)^n\le \lim_{\epsilon\to 0^+}e^{\lambda+\epsilon}=e^\lambda$$
Then we finally can conclude that
$$\lim_{n \rightarrow \infty} \left( 1 + \frac{\lambda}{n}+ o \left( \frac{1}{n} \right) \right)^n = e^{\lambda}$$
