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I want to find

$$\lim_{x\rightarrow0^+}xe^{1/x}$$

This is what my textbook does:

$$\lim_{x\rightarrow0^+}x\cdot e^{1/x}=\left[x=\frac{1}{t}\right]=\lim_{x\rightarrow+\infty}\frac{e^t}{t}=+\infty$$

This is indeed correct. It basically replaces the variable x with $1/t$. What I instead did was:

  • $x \rightarrow0^+$ (I replace x with the value that the limit approaches to, since $x$ is a continuous function).
  • For $e^{1/x}$, this is a composition of function: $\frac 1 x \rightarrow+\infty$ for $x\rightarrow0^+$, therefore $e^{+\infty}=+\infty $ for $x\rightarrow0^+$.

This results in the product of $0^+$ and $+\infty$ which should be equal to $0^+$. I wonder what I did wrong in my reasoning. Any hints?

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  • $\begingroup$ It is wrong to deduce that "zero times infinite must be zero"...In limits, that in fact can be anything between zero and infinity... $\endgroup$ – DonAntonio Aug 8 '18 at 11:10
  • $\begingroup$ @DonAntonio, thanks. So you mean $0^+ \cdot \infty = +\infty$? $\endgroup$ – Cesare Aug 8 '18 at 11:12
  • $\begingroup$ No, I mean (in limits, of course) that $\;0^+\cdot\infty =$ anything in $\;[0,\infty)\;$, depending on each particular case. $\endgroup$ – DonAntonio Aug 8 '18 at 11:14
  • $\begingroup$ Example, $\lim _{x\to 0} x \cdot \frac{1}{x}$. $\endgroup$ – orlp Aug 8 '18 at 11:21
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"This results in the product of $0^+$ and $+\infty$ which should be equal to $0^+$."

I'm afraid you can't do that since $0\cdot\infty$ is undefined. Think of the two examples:

  1. If $x\to\infty$ and $\frac1x\to0$ then $\left(x\cdot\frac1x\right)\to\left(\infty\cdot0\right)=0$ right? But we have $\left(x\cdot\frac1x\right)=1$ so is the limit $0$ or $1$?

  2. If $x^2\to\infty$ and $\frac1x\to0$ then $\left(x^2\cdot\frac1x\right)\to\left(\infty\cdot0\right)=0$ too right? But if $\left(x^2\cdot\frac1x\right)=x$ we should have $\left(x^2\cdot\frac1x\right)=x\to\infty$ so is the limit $0$ or $\infty$?

In essence, you're not allowed to do $0\cdot\infty$ as this is undefined. You have to treat your limits more carefully.

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The $0^+$ is not a real number just means something tends to $0$. So the answer is $\infty$ suggests $e^{\frac{1}{x}}\rightarrow \infty$ is faster than $x\rightarrow 0^+$.

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Your method is wrong because it does not account for rates. Namely, the rate that $a + |b|x$ goes to $\infty$ for $x \to \infty$ is linear, while $e^x$ goes to $\infty$ in an exponential rate that is "faster", hence, $ x e ^x$ will go to $\infty$ much faster than to $0$.

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With your "method" you would get

$ \lim_{x \to 0+}\frac{x}{x}= \lim_{x \to 0+}x\frac{1}{x}=0$, which is false .

Another example: with your "method": $ \lim_{x \to 0+}\frac{x^2}{x}= \lim_{x \to 0+}x^2\frac{1}{x}=0$, which is false !

Consequence: leave $0 \cdot \infty$ undefined !

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As noted we can't conclude in that way since $0\cdot \infty$ is an indeterminate form, to show that limit is $\infty$ we can observe that $\forall t\in \mathbb{R^+}$ we have $e^t>t^2$ and therefore

$$xe^{1/x}\ge x \cdot\frac1{x^2} =\frac1x\to \infty$$

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