MHB Why Doesn't the Limit of \(xe^{-\frac{1}{x}}\) Exist as \(x \to 0\)?

[Vali]

Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?

 

[skeeter]

Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?

the limits from both sides of zero are not equal

$\lim_{x \to 0^+} x \cdot e^{-1/x} = 0$

$\lim_{x \to 0^-} x \cdot e^{-1/x} = -\infty$

 

[HOI]

Why the following limit doesn't exists ?
$$\lim_{x\rightarrow 0}xe^{-\frac{1}{x}}$$
I think it's because of $\frac{1}{x}$ which doesn't exists, right ?

No. As skeeter said, it is because the limits "from below" and "from above" are not the same. If the problem were $$\lim_{x\rightarrow 0^+}xe^{-\frac{1}{x}}$$ it would still be true that "the limit, as x goes to 0, of $\frac{1}{x}$ does not exist" but as $-\frac{1}{x}$ goes to negative infinit, $e^{-\frac{1}{x}}$ goes to 0. $$\lim_{x\rightarrow 0^+}xe^{-\frac{1}{x}}= 0$$.