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

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

No limit as x approaches infinity.

5. anonymous

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

6. anonymous

not much to finding the limit there

7. anonymous

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

8. 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

9. anonymous

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. 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

14. anonymous

sorry, i did not notice you changed the function

15. anonymous

$\lim_{x\rightarrow\infty} x\ln(1-\frac{2}{x})$ so first you notice that you get the indeterminate form $\infty 0$

16. anonymous

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

17. 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

18. anonymous

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