Limit of an exponential sequence 
Show that $$\lim_{n\to\infty}=\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}=-1$$

So what I thought about is filling in the definition of the exponential function, what i get there is $$\lim_{n\to\infty} \frac{\sum_{k=0}^\infty\frac{\left(\frac{-x}{n}\right)^k}{k!}-1}{\frac{-x}{n}}=-1$$
I then eliminate the -1 by adding it to the sum and after that do some index switching, so that i finally get $$\lim_{n\to\infty}(-1)\cdot\sum_{k=0}^\infty\frac{\left(\frac{-x}{n}\right)^k}{(k+1)!}=-1$$
If I then try to solve the limit of the series, i get 0 by Quotient criterium, which is apparently wrong.
Any ideas?
Kind Regards
 A: The limit should be $1$. 
If you want evaluate this limit by using the definition of the exponential function as a power series then, for $x\not=0$,
$$\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}-1=
\frac{\sum_{k=0}^\infty\frac{\left(\frac{-x}{n}\right)^k}{k!}-1}{\frac{-x}{n}}-1=-\frac{x}{n}\sum_{k=2}^\infty\frac{\left(\frac{-x}{n}\right)^{k-2}}{k!}.$$
Now note that
$$\left|\sum_{k=2}^\infty\frac{\left(\frac{-x}{n}\right)^{k-2}}{k!}\right|
\leq \sum_{k=2}^\infty\frac{\left(\frac{|x|}{n}\right)^{k-2}}{(k-2)!}=e^{|x|/n}.$$
Hence, as $n$ goes to infinity,
$$\left|\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}-1\right|\leq 
\frac{|x|e^{|x|/n}}{n}\to 0$$
and we may conclude that
$$\lim_{n\to\infty}\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}=1.$$
A: An alternative, simpler way: note that $$\exp\left(-\frac{x}{n}\right)=1-\frac{x}{n}+\mathcal{O}\left(\frac{x^2}{n^2}\right)$$
truncating the series at the second term.
Thus $$\lim_{n\to\infty}\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}=\lim_{n\to\infty}\frac{1-\frac{x}{n}-1}{\frac{-x}{n}}=1$$
A: One way to find that limit is to use the fact that $$\lim_{x\to 0}\frac{\mathrm{e}^x-1}{x}=1$$
since $\frac{-x}n\to0$ as $n\to\infty$ by setting $t=\frac{-x}n$ your limit becomes $$\lim_{n\to\infty}\frac{\mathrm{e}^{-x/n}-1}{-x/n}=\lim_{t\to0}\frac{\mathrm{e}^t-1}{t}$$
