Find the limit $(\frac{n}{n+5})^n$ 
Find the limit -$$\left(\frac{n}{n+5}\right)^n$$

I set it up all the way to $\dfrac{\left(\dfrac{n+5}{n}\right)}{-\dfrac{1}{n^2}}$ but now I am stuck and do not know what to do.
 A: $$\mathrm L=\lim_{n \to \infty}\left(\frac{n}{n+5}\right)^n \implies \ln {\mathrm L}=\lim_{n \to \infty} n \underbrace{\left(\frac{\ln \left( 1-\frac{5}{n+5} \right)}{\frac{-5}{n+5}} \right)}_{\text{This limit is 1}} \times \frac{-5}{n+5}$$$$ \implies \ln{\mathrm L}=\lim_{n \to \infty}n \cdot \left(\frac{-5}{n+5}\right)=-5$$
$$\implies \mathrm L=e^{-5}$$
A: It's easier to turn it over.
$\left(\frac{n+5}{n}\right)^n
=\left(1+\frac{5}{n}\right)^n
\to e^5
$
so
$\left(\frac{n}{n+5}\right)^n
\to e^{-5}
$.
A: Hint: $\frac{n}{n+5} = 1 + \frac{-5}{n+5} $
A: Just $$\left(1-\frac{5}{n+5}\right)^{-\frac{n+5}{5}\cdot\frac{-5n}{n+5}}\rightarrow e^{-5}$$
A: Alternatively:
$$\lim_\limits{n\to\infty} \left(\frac{n}{n+5}\right)^n=\lim_\limits{n\to\infty} \frac{1}{\left(1+\frac{5}{n}\right)^n}=$$
$$\lim_\limits{n\to\infty} \frac{1}{\left(\underbrace{\left(1+\frac{1}{\frac{n}{5}}\right)^{\frac{n}{5}}}_{=e}\right)^5}=\frac{1}{e^5}.$$
A: Let us assume the fundamental limit $$\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^{n}=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n}=e\tag{1} $$ Taking reciprocals we get $$\lim_{n\to\infty} \left(\frac{n} {n+1}\right)^{n}=\frac{1}{e}\tag{2}$$ And note that the above limit holds if $n$ is replaced by $n+k$ where $k$ is some fixed integer so that $$\lim_{n\to\infty} \left(\frac{n+k} {n+k+1}\right)^{n+k}=\frac{1}{e}$$ and hence $$\lim_{n\to\infty} \left(\frac{n+k} {n+k+1}\right)^{n}=\lim_{n\to\infty}\left(\frac{n+k}{n+k+1}\right)^{n+k}\left(\frac{n+k}{n+k+1}\right)^{-k}=\frac{1}{e}\cdot 1^{-k}=\frac{1}{e}\tag{3}$$ We can now see easily that $$\left(\frac{n} {n+5}\right)^{n}= \left(\frac{n} {n+1}\right)^{n} \left(\frac{n+1} {n+2}\right)^{n} \left(\frac{n+2} {n+3}\right)^{n} \left(\frac{n+3} {n+4}\right)^{n} \left(\frac{n+4} {n+5}\right)^{n} $$ From $(3)$ we can see that each factor on the right of above equation tends to $1/e$ and hence the whole expression tends to $1/e^{5}$.
Using this approach starting from $(1)$ we can easily show via simple algebraic manipulation that $$\lim_{n\to\infty} \left(1+\frac{x}{n} \right) ^{n} =\lim_{n\to\infty} \left(1-\frac{x}{n}\right)^{-n}=e^{x}$$ for rational $x$. 
