Question

lim xln(x^2)
x->oo

Answer 1

\[\lim_{x\rightarrow\infty}x\ln (x^2)\]

Answer 2

or is it \[\lim_{x\rightarrow 0}x\ln (x^2)\]

Answer 3

Yes that's correct. Need help! Not sure what I'm supposed to do next.

Answer 4

No limit as x approaches infinity.

Answer 5

if x is approaching infinity then xln(x^2) is approaching infinity

Answer 6

not much to finding the limit there

Answer 7

No I'm supposed to use L'hospital's rule or something like that.

Answer 8

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

Answer 9

Okay how about lim

Answer 10

i.e. a really large positive number multiplied by another really large number is just another really large positive number

Answer 11

Sorry how bout lim xln(1-2/x)
x->oo

Answer 12

yes, as x approaches infinity the function is ever increasing to infinity, thus that is an infinite limt

Answer 13

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

Answer 14

sorry, i did not notice you changed the function

Answer 15

\[\lim_{x\rightarrow\infty} x\ln(1-\frac{2}{x})\]

so first you notice that you get the indeterminate form \[\infty 0\]

Answer 16

and you need either \[\frac{0}{0}\operatorname{or}\frac{\infty}{\infty}\] to use L'Hopital

Answer 17

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

Answer 18

after one iteration of L'Hopital's rule I get the limit is -2