anonymous
  • anonymous
lim xln(x^2) x->oo
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\lim_{x\rightarrow\infty}x\ln (x^2)\]
anonymous
  • anonymous
or is it \[\lim_{x\rightarrow 0}x\ln (x^2)\]
anonymous
  • anonymous
Yes that's correct. Need help! Not sure what I'm supposed to do next.

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anonymous
  • anonymous
No limit as x approaches infinity.
anonymous
  • anonymous
if x is approaching infinity then xln(x^2) is approaching infinity
anonymous
  • anonymous
not much to finding the limit there
anonymous
  • anonymous
No I'm supposed to use L'hospital's rule or something like that.
anonymous
  • anonymous
if x was approaching zero then you may have to but as x goes to infinity that function goes to infinity, i.e. the form you get \[\infty\infty\] is not indeterminate
anonymous
  • anonymous
Okay how about lim
anonymous
  • anonymous
i.e. a really large positive number multiplied by another really large number is just another really large positive number
anonymous
  • anonymous
Sorry how bout lim xln(1-2/x) x->oo
anonymous
  • anonymous
yes, as x approaches infinity the function is ever increasing to infinity, thus that is an infinite limt
anonymous
  • anonymous
This is probably just a trick question thrown into your problem set where you are using L'Hopital's rule to see if they can catch you off guard
anonymous
  • anonymous
sorry, i did not notice you changed the function
anonymous
  • anonymous
\[\lim_{x\rightarrow\infty} x\ln(1-\frac{2}{x})\] so first you notice that you get the indeterminate form \[\infty 0\]
anonymous
  • anonymous
and you need either \[\frac{0}{0}\operatorname{or}\frac{\infty}{\infty}\] to use L'Hopital
anonymous
  • anonymous
so rewrite the function in the quivelent form \[\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{2}{x})}{\frac{1}{x}}\] and notice now the form is correct for L'Hopital
anonymous
  • anonymous
after one iteration of L'Hopital's rule I get the limit is -2

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