On the justification of the result of a limit utilizing little-o. (quick question)

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{\lambda}{n}+ o \left( \frac{1}{n} \right) \right)^n = e^{\lambda}$$