The limit of \(xe^{-\frac{1}{x}}\) as \(x \to 0\) does not exist due to the differing behavior of the function from the left and right sides of zero. Specifically, \(\lim_{x \to 0^+} xe^{-\frac{1}{x}} = 0\) while \(\lim_{x \to 0^-} xe^{-\frac{1}{x}} = -\infty\). This discrepancy confirms that the overall limit is undefined. The key factor is the behavior of \(e^{-\frac{1}{x}}\) as \(x\) approaches zero from both directions.
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Vali said: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. $$\lim_{x\rightarrow 0^+}xe^{-\frac{1}{x}}= 0$$.Vali said: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 ?