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anonymous
 5 years ago
lim xln(x^2)
x>oo
anonymous
 5 years ago
lim xln(x^2) x>oo

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\rightarrow\infty}x\ln (x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or is it \[\lim_{x\rightarrow 0}x\ln (x^2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes that's correct. Need help! Not sure what I'm supposed to do next.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No limit as x approaches infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if x is approaching infinity then xln(x^2) is approaching infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not much to finding the limit there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No I'm supposed to use L'hospital's rule or something like that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0i.e. a really large positive number multiplied by another really large number is just another really large positive number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry how bout lim xln(12/x) x>oo

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, as x approaches infinity the function is ever increasing to infinity, thus that is an infinite limt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This 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
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, i did not notice you changed the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\rightarrow\infty} x\ln(1\frac{2}{x})\] so first you notice that you get the indeterminate form \[\infty 0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you need either \[\frac{0}{0}\operatorname{or}\frac{\infty}{\infty}\] to use L'Hopital

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0after one iteration of L'Hopital's rule I get the limit is 2
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