Taking the limit of $n(e^{-1/n}-1)$ as $n$ approaches infinity The form is infinity times zero and that is indeterminate which means I need to use L'Hospital's rule, but I have tried to do that but every time I would find another indeterminate form.  How can I use sneaky algebra or sneaky replacements to find the answer?
 A: By the verys definition of derivative, this limit equals $f'(0)$ where $f(x)=e^{-x}$: Note that $\frac{f(0+h)-f(0)}{h}=n(e^{-1/n}-1)$ if $h=\frac1n$.
A: Hint: try expanding $e^{-1/n}$ as a power series: $$e^{-1/n}=1-\frac{1}{n}+\frac{1}{2!} \frac{1}{n^2}+ \dots=1-\frac{1}{n}+\mathcal{O}(n^{-2}) .$$
A: recall $$e^x=1+x+O(x^2)$$as $x\to0$ hence$$e^{-1/n}=1-1/n+O(1/n^2)$$ as $n\to\infty$ and so $$\lim_{n\to\infty}n(e^{-1/n}-1)=\lim_{n\to\infty}n(-1/n+O(1/n^2))=\lim_{n\to\infty}(-1+O(1/n))=-1$$
A: Note that, in general $$\lim_{\alpha\to 0}\frac{x^\alpha-1}{\alpha}=\log x$$
A: L'Hospital's Rule is not my favourite approach, since a computation replaces insight about the behaviour of the function. But one cannot deny its usefulness. We do two L'Hospital's Rule calculations.
Calculation 1: We want to find
$$\lim_{n\to\infty}\frac{e^{-1/n}-1}{1/n}.\tag{1}$$
Let $x=\frac{1}{n}$. As $n\to\infty$, $1/n\to 0^+$. So if 
$$\lim_{x\to 0^+} \frac{e^{-x}-1}{x}\tag{2}$$
exists, then our limit does, and is the same.
Now use L'Hospital's Rule, taking the derivative of the top and bottom of (2), Since 
$$\lim_{x\to 0^+} \frac{-e^{-x}}{1}=-1,$$
our limit exists and is $-1$.
Calculation 2: This time, we deliberately do things in a suboptimal way, by finding
$$\lim_{y\to\infty}\frac{e^{-1/y}-1}{1/y}.$$
Take the derivative of top and bottom. So we want to find
$$\lim_{y\to\infty}\frac{(-1/y^2)(-e^{-1/y})}{(-1/y^2)}.$$
The above simplifies to 
$$\lim_{y\to\infty}(-e^{-1/y}),$$
which is $-1$.
A: You can only use L hospitals rule, if you have 0/0 or inf/inf form.
So, rewrite your expression as (e^[-1/n] - 1)/(1/n) and now use the rule
