anonymous
  • anonymous
Calculate the limit
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\lim_{x \rightarrow 0+} x e^{1/x}\]
anonymous
  • anonymous
and \[\lim_{x \rightarrow 0+} \left( 1 - \cos x \right) \ln x\]
anonymous
  • anonymous
Try a substitution: \[u=\frac{1}{x}\] As \(x\to0^+\), \(u\to\infty\). So, the limit is rewritten as \[\lim_{u\to\infty}\frac{1}{u}e^u\]

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anonymous
  • anonymous
For the first limit, that is.
anonymous
  • anonymous
then I get (e^u)/u then L'hospital i get e^u that is +inf right?
anonymous
  • anonymous
ok, thats right but what about the secound one? :D
anonymous
  • anonymous
Yep, that's correct. Still not sure about the second one
anonymous
  • anonymous
Yes =) P.S. the reason you can use L'Hopital's there is because it's of the form \(\frac{\infty}{\infty}\)
anonymous
  • anonymous
the usual gimmick is to write \[\frac{e^{\frac{1}{x}}}{\frac{1}{x}}\] if the form is indeterminate
anonymous
  • anonymous
yep thats right ! ;D I was assuming that i guess
anonymous
  • anonymous
can probably use that same gimmick for the second one
anonymous
  • anonymous
how so?
anonymous
  • anonymous
I'm wondering if you can write the second limit as \[\lim_{x\to0^+}\frac{\ln x}{\frac{1}{1-\cos x}}\]
anonymous
  • anonymous
yes u can
anonymous
  • anonymous
i did that but the result after L'hospital is the opossite of the real answer
anonymous
  • anonymous
Might be a mistake with your derivative-taking ?
anonymous
  • anonymous
yes u are correct!!!
anonymous
  • anonymous
it was a problem in my derivative taking!
anonymous
  • anonymous
ty for ur time!
anonymous
  • anonymous
You're welcome!
anonymous
  • anonymous
but I have another question that I'll ask in another topic ok? please check it out
anonymous
  • anonymous
Is it not a problem that that limit approaches \(\frac{-\infty}{\infty}\)
anonymous
  • anonymous
@SmoothMath, L'Hopital's rule works with that.
anonymous
  • anonymous
Lovely.

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