Why do we have $(1+\frac{c}{n}+o(\frac{1}{n}))^n \to e^c$ as $n \to \infty?$

Question

We have $(1+\frac{c}{n})^n \to e^c$ as $n \to \infty$. Why does a term in $o(\frac{1}{n})$ does not change this asymptotic?

Answer 1

Suppose we let the $o(1/n)$ term as $g(n)$ ($\in o(1/n)$).

Then by definition, $$\lim_{n\rightarrow\infty} \frac{g(n)}{1/n} = \lim_{n\rightarrow\infty} ng(n) = 0$$

Thus if we look at the limit, $$\lim_{n\rightarrow\infty} (1+\frac{c}{n}+g(n))^n = \lim_{n\rightarrow\infty}\left(1+\frac{c+ng(n)}{n}\right)^{n\cdot \frac{c+ng(n)}{c+ng(n)}} = \lim_{n\rightarrow\infty}\left(1+\frac{c+ng(n)}{n}\right)^{\frac{n}{c+ng(n)}\cdot (c+ng(n))}$$ $$=\exp(\lim_{n\rightarrow\infty}(c+ng(n))) = e^c$$

as desired.

Answer 2

If $x_n=o(\frac 1 n)$ then $n\log(1+\frac c n+x_n)-n\log(1+\frac c n)=n\log(1+\frac {x_n} {1+\frac c n})$. This behaves like $\frac {nx_n} {1+\frac c n}$ for large $n$. [ Because $\log(1+x)$ behave like $x$ for $ x$ near $0$]. Of course $\frac {nx_n} {1+\frac c n} \to 0$. By squeeze theorem we see that $n\log(1+\frac c n+x_n)-n\log(1+\frac c n) \to 0$. Take exponential to complete the proof.