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I just review the following problem:
How to find the limits $\lim\limits_{h\rightarrow 0} \frac{e^{-h}}{-h}$ and $\lim\limits_{h\rightarrow 0} \frac{|\cos h-1|}{h}$?

However, I cannot still know how to solve the following: How to find the following: $$\lim_{x\rightarrow 0^+} \frac{e^{-a/x}}{x}, \ \ a>0$$

By L'hospital's rule:
$$\lim_{x\rightarrow 0^+} \frac{e^{-a/x} \frac{a}{x^2}}{1}= \lim_{x\rightarrow 0^+} \frac{e^{-a/x} a}{x^2}$$

it seems that the degree of the denominator will increase; however, I am still confused about the limit of this problem. Please advise, thanks!

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First note that the derivative approach isn't going to fly, $e^{-a/x}$ is not even continuous at $0$!

Rewrite your limit as $$ \lim_{x\to 0^+}\frac{1/x}{e^{a/x}} $$ note that both top and bottom tend to positive infinity. Apply L'Hopital's, $$ \lim_{x\to 0^+}\frac{1/x}{e^{a/x}}= \lim_{x\to 0^+}\frac{-1/x^2}{-a/x^2e^{a/x}}\\ =\frac1 a\lim_{x\to 0^+}\frac{1}{e^{a/x}}=0 $$

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Note that for $a>0$

$$ \frac{e^{-a/x}}{x}=e^{\log e^{-a/x}-\log x}=e^{-\frac{a}{x}-\log x}\to e^{-\infty}=0$$

indeed

$$-\frac{a}{x}-\log x=\frac1x\left(-a-x\log x\right)\to+\infty\cdot(-a-0)=-\infty$$

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  • $\begingroup$ @MartinR Thanks Martin it was just a typo! $\endgroup$ – gimusi Jan 18 '18 at 8:20
  • $\begingroup$ You could add an argument why $-\frac{a}{x}-\log x \to -\infty$ holds. $\endgroup$ – Martin R Jan 18 '18 at 8:22
  • $\begingroup$ @MartinR of course you are right, I've added this argument, Thanks $\endgroup$ – gimusi Jan 18 '18 at 8:25
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A possible approach without L'Hôpital's rule: Since $$ e^y = \sum_{n=0}^\infty \frac{y^n}{n!} > \frac{y^2}{2} $$ for $y > 0$, we have $$ 0 < \frac{e^{-a/x}}{x} = \frac{1}{xe^{a/x}} < \frac{2x}{a^2} $$ so that the limit is zero. The same argument can be used to show that $$ \lim_{x \to 0^+} \frac{e^{-a/x}}{x^k} = 0 $$ for $a > 0$ and any exponent $k > 0$. The essential idea is that the exponential function "grows faster than any polynomial".

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  • $\begingroup$ very nice proof! $\endgroup$ – gimusi Jan 18 '18 at 8:28
  • $\begingroup$ Martin. Deleted my post. $\endgroup$ – Peter Szilas Jan 18 '18 at 11:51
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We can use the standard result $x^n/e^x\to 0$ as $x\to\infty$. For the current question just put $a/x=t$ so that $t\to\infty$ and the given expression equals $(1/a)(t/e^t)$ which tends to $(1/a)\cdot 0=0$.

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\begin{align*} \lim_{x\rightarrow 0^{+}}\dfrac{e^{-a/x}}{x}&=\lim_{x\rightarrow 0^{+}}\dfrac{x^{-1}}{e^{a/x}}\\ &=\lim_{x\rightarrow 0^{+}}\dfrac{-x^{-2}}{e^{a/x}\cdot-ax^{-2}}\\ &=\lim_{x\rightarrow 0^{+}}\dfrac{1}{a}\dfrac{1}{e^{a/x}}\\ &=\dfrac{1}{a}\cdot 0\\ &=0. \end{align*}

Another way: \begin{align*} \lim_{x\rightarrow 0^{+}}\dfrac{e^{-a/x}}{x}&=\lim_{u\rightarrow\infty}ue^{-au}\\ &=\lim_{u\rightarrow\infty}\dfrac{u}{e^{au}}\\ &=\lim_{u\rightarrow\infty}\dfrac{1}{ae^{au}}\\ &=0. \end{align*}

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