Question

Calculate the limit

Answer 1

\[\lim_{x \rightarrow 0+} x e^{1/x}\]

Answer 2

and
\[\lim_{x \rightarrow 0+} \left( 1 - \cos x \right) \ln x\]

Answer 3

Try a substitution:
\[u=\frac{1}{x}\]
As \(x\to0^+\), \(u\to\infty\). So, the limit is rewritten as
\[\lim_{u\to\infty}\frac{1}{u}e^u\]

Answer 4

For the first limit, that is.

Answer 5

then I get (e^u)/u then L'hospital i get e^u that is +inf right?

Answer 6

ok, thats right but what about the secound one? :D

Answer 7

Yep, that's correct. Still not sure about the second one

Answer 8

Yes =)
P.S. the reason you can use L'Hopital's there is because it's of the form \(\frac{\infty}{\infty}\)

Answer 9

the usual gimmick is to write
\[\frac{e^{\frac{1}{x}}}{\frac{1}{x}}\] if the form is indeterminate

Answer 10

yep thats right ! ;D I was assuming that i guess

Answer 11

can probably use that same gimmick for the second one

Answer 12

how so?

Answer 13

I'm wondering if you can write the second limit as
\[\lim_{x\to0^+}\frac{\ln x}{\frac{1}{1-\cos x}}\]

Answer 14

yes u can

Answer 15

i did that but the result after L'hospital is the opossite of the real answer

Answer 16

Might be a mistake with your derivative-taking ?

Answer 17

yes u are correct!!!

Answer 18

it was a problem in my derivative taking!

Answer 19

ty for ur time!

Answer 20

You're welcome!

Answer 21

but I have another question that I'll ask in another topic ok? please check it out

Answer 22

Is it not a problem that that limit approaches \(\frac{-\infty}{\infty}\)

Answer 23

@SmoothMath, L'Hopital's rule works with that.

Answer 24

Lovely.