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I am trying to study a function defined as $$ f(x)=\begin{cases} x\exp\left(\dfrac{1}{x}\right); & x ≠ 0\\ 0; & x=0 \end{cases} $$ I guess the limit of $f(x)$ as $x → 0$ is undefined ($∞ × 0$) and thus the function would not be continuous at $0$.

But I would like to investigate the left and right handed limits of $f(x)$ to know which of them tends to $0$ so as to know if it is continous from the left or from the right at $0$.

Could someone show me how to investigate the left and right limits of $f(x)$ as $x → 0$?

Ѕᴀᴀᴅ
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Lee
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  • Have you tried setting $t=1/x$? – Hans Lundmark Jun 24 '18 at 15:17
  • Your guess is premature. A limit is there to lift "undefinedness" and assign a value to a $0\times\infty$. You need to evaluate the limit before you can conclude. –  Jul 13 '18 at 18:16

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Substituting $t=1/x$, we find that $$\lim_{x\to 0,\,x>0}\,x\exp(1/x)=\lim_{t\to+\infty}\frac{e^t}{t}=+\infty$$ and $$\lim_{x\to 0,\,x<0}\,x\exp(1/x)=\lim_{t\to-\infty}\frac{e^t}{t}=0.$$ Note that when $x<0$, $t$ approaches $-\infty$ instead of $+\infty$ as $x$ approaches $0$, since $1/x$ is negative in this case. Thus, if we restrict the domain of the function to $x\leq0$, it becomes continuous.

apanpapan3
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  • apanpapan3, thanks for commenting, but why did you delete your comment in https://math.stackexchange.com/questions/2867643 ? – BCLC Jul 31 '18 at 05:44