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
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 $$\lim_{n\to\infty}\frac{\exp^\left(\frac{-x}{n}\right)-1}{\frac{-x}{n}}=1.$$