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?

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
