-2
$\begingroup$

The equality in question is widely used to solve for limits like $\lim_{n\to\infty} \frac{n+1}{n-2}$, but how do you actually prove that
$$\lim_{n\to\infty} \left(1+\frac{k}{n}\right)^n = e^k.$$ Is it possible to prove it using the other formula: $\lim(1+\frac 1 n)^n = e?$

EDIT: I think I already got it: $$(1+ \frac{k}{n})^n = (1+ \frac{1}{\frac{n}{k}})^{k \cdot \frac{n}{k}} \rightarrow e^k$$ Is this correct?

$\endgroup$

marked as duplicate by Sil, MrYouMath, Guy Fsone, Math Lover, Mark Bennet Nov 21 '17 at 18:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

Since $f(x)=x^k$ is a continuous function, we obtain:

$$\lim_{n\to\infty}\left(1 +\frac k n\right)^n = \lim_{n\to\infty}\left(1 +\frac k n\right)^{\frac{n}{k}\cdot k } = \left(\lim_{n\to\infty}\left(1 +\frac k n\right)^{\frac{n}{k} }\right)^k=e^k$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.