Use L'Hopital's rule to evaluate the limit

In summary, to evaluate the limit using L'Hopital's rule, the conversation discussed using logarithms and taking derivatives. However, the derivatives did not help in solving the limit, so it was suggested to use L'Hopital's rule on the logarithm expression instead.
  • #1
physstudent1
270
1

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
 
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  • #2
my guess would be you have to use some log properties like log(a/b) = log(a) - log(b)
 
  • #3
i did do that to get to the derivative but the derivatives didn't help
 
  • #4
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)
 
  • #5
Why use L'Hopital? Just look at its reciprocal function. Trivial.
 
  • #6
Kummer said:
Why use L'Hopital?

To get more experience using it?
 
  • #7
physstudent1 said:

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

What is L'Hopital's rule?

L'Hopital's rule is a mathematical tool used to evaluate limits of indeterminate forms, where the limit of the numerator and denominator approach either 0 or infinity. It states that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives.

When should I use L'Hopital's rule to evaluate a limit?

L'Hopital's rule should only be used when the limit of the function is in an indeterminate form, such as 0/0 or ∞/∞. It is not applicable to all limits and should be used with caution.

How do I use L'Hopital's rule to evaluate a limit?

To use L'Hopital's rule, take the derivative of both the numerator and denominator of the function. Then, evaluate the limit of the new function. If it is still indeterminate, repeat the process until the limit can be determined.

Can L'Hopital's rule be used for all types of functions?

No, L'Hopital's rule can only be used for functions that are differentiable in a given interval and have a limit in that interval. It cannot be used for discontinuous functions or functions with infinite oscillations.

Are there any common mistakes when using L'Hopital's rule?

Yes, some common mistakes include using the rule when it is not applicable, taking the derivative incorrectly, and using it to evaluate limits at infinity. It is important to carefully check the conditions for using L'Hopital's rule and to double-check the derivative calculations.

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