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How to compute the limits for the following functions without using L'Hopital rule?

1) $\displaystyle\underset{x\to 0^{+}}{\lim} \frac{e^{1/x} + 2 e ^{-1/x} + \ln x}{3e^{1/x} + 5e^{-1/x} + 7\ln x} $

2) $\displaystyle\underset{x\to 0^{+}}{\lim} \frac{x+e^{-1/x}}{x-e^{-1/x}}$

I wouldn't know where to begin. Thank you very much.

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    $\begingroup$ Hint: divide the numerator and denominator by the dominating term. i.e, for the first one divide through by $e^{1/x}$. Also, is there a typo in the second one? should the exponential in the denominator be $e^{-1/x}$? $\endgroup$ – Leon Sot Jan 20 '17 at 10:11
  • $\begingroup$ Thank you for the hint. I will try it out now. Yes, that was a typo, and I have edited it now. $\endgroup$ – ForumWhiner Jan 20 '17 at 10:19
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    $\begingroup$ i personally find it more convenient to subsitute $y=1/x$ and inspecting the limits at infinity for this kind of problems. So for 2) $$ \lim_{y\rightarrow \infty} \frac{1+ye^{-y}}{1-ye^{-y}} $$ Now since $ye^{-y}<1$ on $(0,\infty)$we can apply the geometric series. Furthermore since $ye^{-y}\rightarrow 0$ as $y\rightarrow \infty$ we easliy conclude that the limit in question is $1$ $\endgroup$ – tired Jan 20 '17 at 10:20
  • $\begingroup$ I see, thank you very much. $\endgroup$ – ForumWhiner Jan 20 '17 at 10:28
  • $\begingroup$ Any further hints on the first problem? I have shown below in answer section how far I got (not far by the way!) $\endgroup$ – ForumWhiner Jan 20 '17 at 10:35
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For problem 1, I did this:

Take out $e^{1/x}$

$$\lim_{x\to 0} \frac{1+ 2\,e^{-2/x}+e^{-1/x}\ln x}{3 + 5\,e^{-2/x} + 7\,e^{-1/x}\ln x}$$

Now what?

For problem 2, I did this:

Substitute $y=1/x$

$$\lim_{x\to 0} \frac{1/y + e^{-y}}{1/y - e^{-y}} $$

Now what?

Thank you very much.

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  • $\begingroup$ For the first one, the $e^{-2/x}$ go to zero and so you just need to show that $e^{-1/x}\ln x$ also goes to zero. To see this you can do the $y=1/x$ substitution and rewrite it as $\frac{-\ln(y)}{e^{y}}$ and this is easily seen to go to zero as $y\to \infty$. $\endgroup$ – Leon Sot Jan 20 '17 at 10:37
  • $\begingroup$ Thank you. So 1/3 remains and this is the answer right? $\endgroup$ – ForumWhiner Jan 20 '17 at 10:41
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Following the hint for 1) you get

$$ \require{cancel} \lim_{x\to 0^{+}} \frac{\cancel{e^{1/x}}}{\cancel{e^{1/x}}}\frac{1 + 2e^{-2/x} + e^{-1/x}\ln x}{3 + 5e^{-2/x} + 7e^{-1/x}\ln x} $$

And then we observe that, for $x\to 0$, $e^{-1/x}$ wins over $\ln x$

$$ \left\vert \frac{\ln x}{e^{1/x}} \right\vert \leq \left\vert \frac{x-1}{e^{1/x}} \right\vert \underset{x\to0}{\longrightarrow} 0 $$

And I used a well-known logarithm inequality. So the limit is evaluates to

$$ \lim_{x\to 0^{+}} \frac{1 + 2e^{-2/x} + e^{-1/x}\ln x}{3 + 5e^{-2/x} + 7e^{-1/x}\ln x} = \frac{1}{3} $$

Because $e^{-2/x}\to 0$ for $x\to 0$.

For problem 2), following from your answer, do the same trick: get $y$ out of the fraction, and then

$$ \frac{e^{-y}}{y}\underset{y\to\infty}{\longrightarrow} 0 $$

(This is well-known, no inequalities involved) So that the limit for 2) is 1.

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  • $\begingroup$ @egreg Thank you, fixed now $\endgroup$ – Eugenio Jan 21 '17 at 8:43

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