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Use L'Hopital's rule to evaluate the limit

270
1
1. Homework Statement

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

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

2. Homework Equations



3. 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
 
Last edited:
452
0
my guess would be you have to use some log properties like log(a/b) = log(a) - log(b)
 
270
1
i did do that to get to the derivative but the derivatives didn't help
 
452
0
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)
 
291
0
Why use L'Hopital? Just look at its reciprocal function. Trivial.
 

HallsofIvy

Science Advisor
Homework Helper
41,712
876
1. Homework Statement

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

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

2. Homework Equations



3. 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
[tex]\frac{ln(x)- ln(x+1)}{x^{-1}}[/tex]
and apply L'Hopital's rule to that.
 

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