$$\lim\limits_{x\to 0^-} x^3e^{1/x}$$
If I plug in zero, I get $0\cdot\infty$. So, this, I thought was a hint to try to rewrite the problem to try to get $\frac{\infty}{\infty}$ or $\frac{0}{0} so that I could apply L'Hôpital's.
Once I did this and applied L'Hôpital's, I still kept getting $\frac{\infty}{\infty}$.
*** Then, someone in my class pointed out that $$\lim_{x\to 0^-} e^{1/x}=0$$ I checked on the calculator and verified this. But, then, why is it that when you plug in $0$ you get $e^{\infty}$ which is infinity? If I were to run into this question on a test where no calculator was allowed, I would've thought this to be $e^{\infty}$. Can someone explain this?
***Omg I think I just found out the answer to my own question....(well actually, Google helped me) but I just want to make sure...is it because you're plugging in NEGATIVE very small numbers into x...so then it would be $e^{-\infty}$ which is basically zero. Yes or no?
There is another way to compute this limit. We can write $$ \lim_{x\to 0^-}x^3e^{\frac{1}{x}}=\lim_{x\to 0^-}x^3\lim_{x\to 0^-}e^{\frac{1}{x}}. $$ The first limit quite obviously goes to zero. For the second limit, let $\epsilon>0$ and $\delta=\frac{1}{\epsilon}.$ Then $$ -\frac{1}{x}<-\frac{1}{1/\epsilon}=-\epsilon~\forall ~x\in (-\delta,0). $$ Since we can choose $\epsilon$ arbitrarily large, conclude $$ \lim_{x\to 0^-}\frac{1}{x}=-\infty . $$ Hence we have $$ 0\times e^{-\infty}=0. $$