
#1
Oct605, 05:00 PM

P: 341

Hi. Use L'Hopital's Rule to evaluate the limit.
lim xinfinity of (lnx) ^(2/x) The answer is 1. I kept taking the derivative but it seemed like I was going around in circles. Any help would be appreciated. Thanks 



#2
Oct605, 05:03 PM

HW Helper
P: 402

L'Hopital's Rule does not apply to the function as written. What were your first steps?




#3
Oct605, 05:08 PM

P: 341

I found the derivative of the function and it was really ugly. I don't know if it is correct.
[ [2(lnx)^(2/x 1)][lnx*ln(lnx) 1] ]/x^2 



#4
Oct605, 05:11 PM

HW Helper
P: 1,024

Use L'Hopital's Rule to evaluate the limit
As Diane said, you can't apply L'Hopital's rule to this function, at least not in this form.
It is possible however to alter the function so it becomes something where you can use L'Hopital. You need to convert it to an undeterminate form of 0/0 or inf/inf. 



#5
Oct605, 05:12 PM

P: 341

How can I convert it?




#6
Oct605, 05:33 PM

HW Helper
P: 1,024

Well, what you have here is a case of f(x)^g(x) which yields the indeterminate form [itex]\infty ^0[/itex].
You can convert it to another indeterminate form by doing [itex]\exp \left( {\ln \left( {f\left( x \right)^{g\left( x \right)} } \right)} \right) = \exp \left( {g\left( x \right)\ln \left( {f\left( x \right)} \right)} \right)[/itex]. Then you have something of the form f(x)g(x) which gives a new indeterminate form [itex]\infty \cdot 0[/itex]. Finally, you can convert this to f(x)/(1/g(x)) (or the other way arround) to get either 0/0 or inf/inf so that you can use L'Hopital. 


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