0
$\begingroup$

I found a limits equation

$$\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^n=e^{-\lambda}$$

How can I get the result of $e^{-\lambda}$?

Normally, we can use

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

And how can I get $e^n$?

$\endgroup$
  • $\begingroup$ you must stablish a relation between $\left(1+\frac{y}{x}\right)^x$ and $\left(1+\frac1{x}\right)^{xy}$ $\endgroup$ – Masacroso Aug 2 '17 at 9:07
2
$\begingroup$

You may know that (sometimes this is used as definition of $e$) $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e $$ Taking $k$th powers, $k\in\Bbb N$, we obtain $$e^k=\lim_{n\to\infty}\left(1+\frac1{n}\right)^{nk}=\lim_{n\to\infty}\left(1+\frac k{nk}\right)^{nk}.$$ The latter limit is the limit of a subsequence of $\lim_{n\to\infty}\left(1+\frac k{n}\right)^{n}$, hence this also converges to $e^k$, once we know it converges at all. In fact, the same method shows that more generally $$\lim_{n\to\infty}\left(1+\frac {ak}n\right)^n =\left(\lim_{n\to\infty}\left(1+\frac {a}n\right)^n\right)^k$$ for $k\in\Bbb N$ and arbitrary $a$ (provided both limits exist). As a consequence, $$\lim_{n\to\infty}\left(1+\frac {a}n\right)^n=e^a\qquad \text{for all }a\in\Bbb Q_{\ge0}.$$ Finally, using $(1-\frac1n)^n(1+\frac1n)^n=(1-\frac1{n^2})^n$, you can show that the same also hods for $a=-1$ and hence also for all $a\in\Bbb Q$.

$\endgroup$
0
$\begingroup$

It is actually this, lambda means here a variable.$$\lim_{n \to \infty}\left(1-\frac{x}{n}\right)^n=e^{-x}$$

So, for example

$$\lim_{n \to \infty}\left(1-\frac{5}{n}\right)^n=e^{-5}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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