Question

LHospital Rule lim x->infinity (1=(2/x^2))^lnx

Answer 1

I can't make sense of this

Answer 2

nor me!

Answer 3

\[\lim x-> \infty (1+2/x ^{2})^{lnx}\]

Answer 4

as x --> inf 2 /x^ .---> 0
and 1 to any power = 1

limit is 1

Answer 5

1 is the answer...but your method doesn't always work

Answer 6

Using L'hopital you should first get to one of the forms allowed from the rule. What you can do is apply a function and its inverse so that everything remains unchanged but it gives you the opportunity to go to a form such as infinity/infinity.
I usually use the exponential and the logarithm and then use them properties.

Answer 7

woops... their* . English mistake :P

Answer 8

please explain zarkon

Answer 9

@ jimmyrep
compute this limit

\[\lim_{x\to\infty}\left(1+\frac{2}{x^2}\right)^{x^2}\]

Answer 10

I'll give you a hint...the answer is not 1

Answer 11

you've got me there - limits are'nt my strong point!

Answer 12

For the original problem--but not Zarkon's, which is very important-- consider the log of the expression:
\[ \ln (1+2/x^2)^{\ln x} = \ln x . \ln(1 + 2/x^2) = \frac{\ln(1 + 2/x^2)}{ 1/\ln x} \]

Answer 13

Now you can apply l'Hopital's rule.

Answer 14

aahh zarkons problem its the exponential! limit is e^x right?

Answer 15

Yes,
\[ \lim_{x \rightarrow \infty} (1 + a/x)^{1/x} = e^a \]

Answer 16

thanks for confirming this - i a[ways have trouble with limits. to me they are most difficult part of calculus