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How to prove that

$\lim_{n\rightarrow\infty} (1-\frac{k}{n})^n = e^{-k}$

and

$\lim_{n\rightarrow\infty} (1-\frac{k}{n})^k = 1$?

Any answer will be appreciated. Thanks.

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    $\begingroup$ Hint: Use the Binomial expansion for $(1-x)^{n}$. $\endgroup$ – Sudarsan Sep 23 '13 at 19:08
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    $\begingroup$ For the second: the exponent does not depend on the limit and $$x\mapsto x^k$$ is continuous, so swap $\cdot^k$ and $\lim$. $\endgroup$ – AlexR Sep 23 '13 at 19:12
  • $\begingroup$ not clear, any body can give the simple illustrate? $\endgroup$ – user95984 Sep 23 '13 at 19:22
  • $\begingroup$ The Maple command $$Student[Calculus1]:-LimitTutor((1-k/n)^n, n = infinity); $$ does the job step by step with explanations. See this for info. $\endgroup$ – user64494 Sep 23 '13 at 19:25
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$$ \begin{array}{rclcl} \left(1 - {k \over n}\right)^{n} & = & {\rm e}^{n\ln\left(1\ -\ k/n\right)} = {\rm e}^{n\left(-\ k/n\ -\ k^{2}/2n^{2} - \cdots\right)} & \to & {\rm e}^{-k} \\[3mm] \left(1 - {k \over n}\right)^{k} & = & {\rm e}^{k\ln\left(1\ -\ k/n\right)} = {\rm e}^{k\left(-\ k/n\ -\ k^{2}/2n^{2} - \cdots\right)} & \to & 1 \end{array} $$

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  • $\begingroup$ How is the $ln(1 - k/n)$ transformed into that expansion? Could you include another term of that expansion? $\endgroup$ – user3105485 Mar 5 '17 at 18:54
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    $\begingroup$ @user3105485 Use $z = \exp\left(\ln\left(z\right)\right)$. You can include more terms but thay don't give any contributions in the cited limits. $\endgroup$ – Felix Marin Mar 5 '17 at 21:21
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for the first asks $$\lim_{n \to \infty}{\left(1-\frac{k}{n}\right)}^n=e^{\lim_{n \to \infty}n\ln{\left(1-\frac{k}{n}\right)}}.$$ Now, making the change of variables $\frac{1}{n}=t,$ then

$$e^{\lim_{n \to \infty}n\ln{\left(1-\frac{k}{n}\right)}}=e^{\lim_{t \to 0}\left[\frac{1}{t}\ln{\left(1-kt\right)}\right]}.$$

We have a indeterminate from of type $0/0$. Now, we can apply the L'Hospital's rule and gain $$e^{\lim_{t \to 0}\frac{\ln{\left(1-kt\right)}}{t}}=e^{\lim_{t\to 0}\frac{\frac{d}{dt}\ln(1-kt)}{\frac{d}{dt}t}}=e^{\lim_{t\to 0}\frac{1}{1-kt}(-k)} =e^{-k}.$$

We can use the same approach to solve the second issue..

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It is a fact that $$\lim_{x \to \infty}{\left(1+\frac{1}{x}\right)}^x=e$$ You can find a proof of that here: http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf

Now, $$\lim_{n\to\infty}{\left(1-\frac{k}{n}\right)}^n=\lim_{n\to\infty}{\left[1+\left(-\frac{k}{n}\right)\right]}^n$$ $$=\left[\lim_{n\to\infty}{\left[1+\left(-\frac{k}{n}\right)\right]}^{(-{n}/{k})}\right]^{-k}=e^{-k}$$

For the second limit, k is just a number, it does not tend to infinity, so the previous formula doesnt apply anymore. So, because $\frac{1}{n}\to 0$, we have $$\lim_{n\to\infty}\left(1-\frac{k}{n}\right)^k=1^k=1$$

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