amyna one year ago ok so for this limit it is an infinity over an infinity. But how do i use l hospitals rule for this since its not in fraction form?

1. amyna

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2. phi

you could find the limit of the ln of that expression then the limit of the original will be e^(limit)

3. amyna

so i could do 2x ln x ?

4. phi

yes and write that as $2\frac{\ln x}{x^{-1}}$

5. amyna

ok thats all i needed to know! Thank You!

6. dan815

probably a better idea to do 2x/(1/lnx)

7. amyna

ya that seems much easier

8. dan815

the other way simplfies better :)

9. amyna

i got 0 so do i do the l hospitals rule again?

10. dan815

ya thats right now apply e^0= 1

11. amyna

alrighty!!! Thanks!