Use L'Hopital's rule to evaluate the limit

[physstudent1]

Homework Statement

Use L'Hopital's rule to evaluate the limit:

lim (x/(x+1))^x
x>infinite

Homework Equations

The Attempt at a Solution

I put it into a logarithm first to make it the limit of xln(x/(x+1)) then I took the derivative and got [(1/x)-(1/(x+1))]/[x^-2]
but its still indeterminate and after taking the 2nd derivative it was as well

 

[cse63146]

my guess would be you have to use some log properties like log(a/b) = log(a) - log(b)

 

[physstudent1]

i did do that to get to the derivative but the derivatives didn't help

 

[cse63146]

You forgot to use the power rule since there are 2 functions

This is what I got (could be wrong)

= x(logx - log (x+1)
= logx - log(x+1) + x((1/x)-(1/x-1)) <---simplify
= logx - log(x+1) - (1/x-1)

I believe you need to take the second derivative and you should get the limit to be 0 (it's been about a year since I did diffrential calc)

 

[Kummer]

Why use L'Hopital? Just look at its reciprocal function. Trivial.

 

[cse63146]

Why use L'Hopital?

To get more experience using it?

 

[HallsofIvy]

Homework Statement

Use L'Hopital's rule to evaluate the limit:

lim (x/(x+1))^x
x>infinite

Homework Equations

The Attempt at a Solution

I put it into a logarithm first to make it the limit of xln(x/(x+1)) then I took the derivative and got [(1/x)-(1/(x+1))]/[x^-2]
but its still indeterminate and after taking the 2nd derivative it was as well

Why the derivative? L'Hospital's rule usually involves taking two derivatives separately! Write the logarithm as
\frac{ln(x)- ln(x+1)}{x^{-1}}
and apply L'Hopital's rule to that.