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I need to find lateral limits of this one $$\lim_{x \to 0}xe^{\frac{1}{x}}$$ I tried and I got that when $x$ is smaller than $0$ the limit is $0$. But what do I do when $x$ is bigger than $0$?

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    Could you show how the limit goes to zero when $x<0$? – 2017-02-28
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    You are saying that $e^{1/0} = 1$? – 2017-02-28
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    i am saying that when x is smaller than 0 you get $e^{1/0-}=1/e^\infty=0$ – 2017-02-28
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    Ok then I get it now – 2017-02-28

3 Answers 3

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Is it allowed to use Taylor series in this question? If so, then we have $$\lim_{x\rightarrow0^+}x\cdot e^{\frac1x}\qquad\qquad\qquad\qquad$$ $$\quad=\ \lim_{x\rightarrow0^+}x\cdot\left(1+\frac1{x}+\frac1{2!x^2}+\frac1{3!x^3}+\cdots\right)$$ $$=\ \lim_{x\rightarrow0^+}\left(x+1+\frac1{2!x}+\frac1{3!x^2}+\cdots\right)\ \ \ \,$$ $$=\ +\infty\qquad\qquad\qquad\qquad\qquad\quad\ \ $$

This is true because every fraction with $x$ in its denominator tends to infinity.

Another approach that only involves differentiation and a little analysis:

$\left.\right.$

Let $\,f(y)=e^y,\ g(y)=(y^2)/2$, then we have $$f'(y)=e^y\,\ \text{and}\ \,g'(y)=y$$ Now differentiate again and we get $$f''(y)=e^y\,\ \text{and}\ \,g''(y)=1$$ When $\,y>0$, $\ e^y>1\,$ and that means $\,f''(y)>g''(y)$.

Also, $\ f'(0)=1>0=g'(0)$, so $\,f'(y)>g'(y)\,$ for all $\,y>0$

Again, since $\ f(0)=1>0=g(0)$, so $\,f(y)>g(y)\,$ for all $\,y>0$

So far we have proved that if $\,y>0\,$ then $\,e^y>(y^2)/2$

$\left.\right.$

Now back to the limit, and we have that

$$\lim_{x\rightarrow0}xe^{\frac1x}=\lim_{y\rightarrow\infty}\frac{e^y}y\geq\lim_{y\rightarrow\infty}\frac{(y^2)/2}y=\infty$$

And that is the "exponential dominating a polynomial" in Zahbaz's answer

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    I have heard of Taylor series but I do not know how to use it :( – 2017-02-28
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    I see, so we need to use some very basic knowledge here, and that could be a little tricky then... – 2017-02-28
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    By the way, could you prove $e^x\geq (x^2)/2$ for $x>0$ by comparing their derivatives? If you have proved that, then you can work out the limit – 2017-02-28
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Are you permitted to rely on an exponential dominating a polynomial in the limit? These limits may look more approachable by transforming $x\to \frac{1}{y}$ such that as $x\to0^\pm$, $y\to\pm\infty$.

$$\lim_{x \to 0^+}xe^{\frac{1}{x}}=\lim_{y \to \infty}\frac{e^{y}}{y} = \infty$$

$$\lim_{x \to 0^-}xe^{\frac{1}{x}}=\lim_{y \to -\infty}\frac{e^{y}}{y} = 0$$

The last step would be justified with a Taylor expansion, but I'm unclear if that is in your toolkit. You could use an argument like this for the first limit,

$$\frac{e^{y}}{y} > \frac{e^y}{e^{y/2}}$$

so

$$\lim_{y \to \infty} \frac{e^{y}}{y} > \lim_{y \to \infty}\frac{e^y}{e^{y/2}} = \lim_{y \to \infty}e^{y/2}=\infty$$

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$\lim_{x\to 0+}^{1/}= 0 \cdot \infty$ which is an indefinite form.

We could apply L'Hopital Rule but it can only be done if we have a $0/0$ or $\infty/\infty$ indefinite forms. So rewrite $\lim_{\to0+}^{1/}$ as $$ \lim_{\to 0+}\frac{^{1/}}{\frac{1}{x}} $$ by putting $x$ in the denominator as $1/x$. Then plug in the limit as $\to 0+$, we get the $\infty/\infty$ form. Now use L'Hopital's Rule.

  1. Take the derivatives of numerator and denominator. We get
    $$ \lim_{\to 0+}\frac{-\frac{e^{1/x}}{x^2}}{-\frac{1}{x^2}}\text{ the negatives and $x^2$ cancel} $$
  2. Simplify, then plug in the limit as $\to0+$. We get $\lim_{\to0+} e^{1/x}=\infty$!!!!

A tip: graph the numerator and denominator separately so you can see the trend