Indeterminate Forms / L’Hospitals Rule: x^(2/x)

[hansellitis]

Indeterminate Forms / L’Hospitals Rule: x^(2/x)

lim x^(2/x)
x-->0+

The form is 0^(infinity), so:

ln y = 2/x lnx

=x*lnx/2

The form is (-infinity)/0.

Now this is where I get lost. What step should I take next?

Thanks.

 

[royhaas]

Write ln(x)/x as -(1/x)ln(1/x) to show that this limit is -infinity.

 

[hansellitis]

Write ln(x)/x as -(1/x)ln(1/x) to show that this limit is -infinity.

-(1/x)ln(1/x)

Even if I did that. -(1/x) and ln(1/x) wouldn't exist, because the denominator would equal 0...

 

[stapel]

Even if I did that. -(1/x) and ln(1/x) wouldn't exist, because the denominator would equal 0...

"Taking the limit" is not the same as "evaluating". :shock:

You are not being asked to evaluate the expression at x = 0; you are being asked to find the limit as x tends toward zero (gets very close to zero) from the right. So try taking that limit!

Eliz.

 

[hansellitis]

Even if I did that. -(1/x) and ln(1/x) wouldn't exist, because the denominator would equal 0...

"Taking the limit" is not the same as "evaluating". :shock:

You are not being asked to evaluate the expression at x = 0; you are being asked to find the limit as x tends toward zero (gets very close to zero) from the right. So try taking that limit!

Eliz.

still though..

lim x^(2/x)
x-->0+

=(2*lnx)/x has the form -infinity/0, and the denominator approaches 0, so we need to differentiate again, right?

which gives us 2/x So we need to take the differentiate once more since it has the indeterminate type 0/0.

And we get 0/1 which is 0.

So

lim x^(2/x) = lim e^0 = 1
x-->0+ x-->0+

Is that right? Because in the graph, it appears x^(2/x) approaches 0.

What did I do wrong?

 

[skeeter]

still though..

lim x^(2/x)
x-->0+

=(2*lnx)/x has the form -infinity/0, and the denominator approaches 0, so we need to differentiate again, right?

no ... if you get -inf/0, the limit is -inf ... you can't use L'Hopital again.

However, there is an out ... note that lny -> -inf, so, what does y approach?

 

[hansellitis]

still though..

lim x^(2/x)
x-->0+

=(2*lnx)/x has the form -infinity/0, and the denominator approaches 0, so we need to differentiate again, right?

no ... if you get -inf/0, the limit is -inf ... you can't use L'Hopital again.

However, there is an out ... note that lny -> -inf, so, what does y approach?

e^-inf = 0

ahhhh thanks. But as the denominator approaches 0, doesn't x approach 0? Making it a non real answer?

 

[skeeter]

x is approaching 0 from the right ... correct?

 

[hansellitis]

x is approaching 0 from the right ... correct?

Oh my. haha, I'm so dumb.

So as x approaches 0 from the right, dividing by small values of x increases the quotient (2 lnx)/x.

Thanks for your help, my brain wasn't functioning correctly.