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?

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 $-1$.
