L'Hospital's Rule Homework: Solving lim x→∞

  • Thread starter redsox5
  • Start date
I get[ln(1+1/x)-1/x]/(-1/x^2). That's 0/0, not 1/0.but that's what I got also, but 0/0 is still indeterminant and the answer should be 1In summary, the conversation discusses the use of L'Hospital's rule to solve the limit of x[(x+1)ln(1+(1/x))-1] as x approaches infinity. The participants explore different approaches, including using the chain rule and Taylor expansion. After careful algebraic manipulation and multiple applications of L'Hospital's rule, the final result is found to be 1.
  • #1
redsox5
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0

Homework Statement



lim[tex]_{x -> infin}[/tex] x[(x+1)ln(1+(1/x))-1]



The Attempt at a Solution



Am I able to start this with the chain rule. I'm not sure how to start this.
 
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  • #2
i've been simplifying your original problem to see what form of L'Hospital's you have

[tex]\lim_{x \rightarrow \infty} x[(x+1)\ln(1+\frac{1}{x})-1][/tex]

also is this correct?
 
  • #3
well that's the oringal problem
Is it an infinity times infinity problem, but to use l'hospital's rule it has to be a fraction
 
  • #4
can i leave x on top and move [(x+1)ln(1+(1/x))-1] to the bottom by raising it to the -1 power? Then use l'hospital's rule
 
  • #5
redsox5 said:

Homework Statement



lim[tex]_{x -> infin}[/tex] x[(x+1)ln(1+(1/x))-1]



The Attempt at a Solution



Am I able to start this with the chain rule. I'm not sure how to start this.

I don't know if this is going to be useful to you because I don;t know if you have seen that trick but I can get the answer quickly by using a Taylor expansion of [tex] ln(1+\epsilon) \approx 1 + \epsilon [/tex] for small epsilon. Then simple algebra leads dircetly to the answer.
 
  • #6
ok think i may have a solution

lim x-> infin [tex]\frac{x}{[(x+1)ln(1+(1/x))-1]^-1}[/tex]

then use l'hospital's rule and get

[tex]\frac{1}{-[(x+1)ln(1+(1/x))-1]^-1 * [ln(1+(1/x)+(x+1)(1/1+(1/x))}[/tex]
 
  • #7
yea, I'm not reallu sure about that. I'm not sure if it will be acceptable
 
  • #8
nrqed said:
I don't know if this is going to be useful to you because I don;t know if you have seen that trick but I can get the answer quickly by using a Taylor expansion of [tex] ln(1+\epsilon) \approx 1 + \epsilon [/tex] for small epsilon. Then simple algebra leads dircetly to the answer.

Be careful here, you may be quickly lead to the wrong answer. You need to keep the quadratic term in the expansion as well. I got burnt by that once.
 
  • #9
you mean in my use of the chain rule?..is this the right approach though?
 
  • #10
redsox5 said:
ok think i may have a solution

lim x-> infin [tex]\frac{x}{[(x+1)ln(1+(1/x))-1]^-1}[/tex]

then use l'hospital's rule and get

[tex]\frac{1}{-[(x+1)ln(1+(1/x))-1]^-1 * [ln(1+(1/x)+(x+1)(1/1+(1/x))}[/tex]

This limit isn't even the same as the one you started out with. Write it as
[(x+1)*ln(1+1/x)-1]/(1/x). Now it's 0/0 (though that's not completely clear until you show (x+1)*ln(1+1/x) -> 1. You can use l'Hopital but you've got to be careful.
 
  • #11
so after you use l'hospital's rule the first time you get

[tex]\frac{ln(1+\frac{1}{x})+(x+1)(\frac{1}{1+\frac{1}{x}})}{\frac{-1}{x^{2}}}[/tex]

and that's infin/0 ...right?
 
  • #12
You were the one who brought up the chain rule. The derivative of ln(1+1/x) is [1/(1+1/x)]*(-1/x^2). It's 0/0. Do some algebra on that before you differentiate again.
 
  • #13
wait..i the correct answer -1 since it equals infin/neg infin
 
  • #14
redsox5 said:
wait..i the correct answer -1 since it equals infin/neg infin

infin/neg infin is NOT -1. It's indeterminant. You need to differentiate again. In the end you will find a perfectly finite limit.
 
  • #15
how many time do you have to use l'hospital's rule..it's getting really messy
 
  • #16
after the second time i use it i have 1/0
 
  • #17
redsox5 said:
how many time do you have to use l'hospital's rule..it's getting really messy

This is the last time. But clean up the algebra in the numerator before you differentiate again. Better to do it before than after.
 
  • #18
but it's 1/0 can I still use l'hospitals'?
 
  • #19
redsox5 said:
but it's 1/0 can I still use l'hospitals'?

I get [ln(1+1/x)-1/x]/(-1/x^2) after I do the algebra from the first differentiation. That's 0/0, not 1/0.
 
  • #20
your numerator is wrong, her's the oringal again

lim x-> infin x[(x+1)ln(1+(1/x))-1]
 
  • #21
nrqed said:
I don't know if this is going to be useful to you because I don;t know if you have seen that trick but I can get the answer quickly by using a Taylor expansion of [tex] ln(1+\epsilon) \approx 1 + \epsilon [/tex] for small epsilon. Then simple algebra leads dircetly to the answer.

Tiny correction, there is no constant term in that. So its just epsilon, though as Dick said, we should keep the quadratic term in there as well.
 
  • #22
redsox5 said:
your numerator is wrong, her's the oringal again

lim x-> infin x[(x+1)ln(1+(1/x))-1]

I don't need the original. That's at the top of the page. I would like to know what you think the result of the first application of l'Hopital's rule is.
 
Last edited:

What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical rule that helps to evaluate limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of a quotient of two functions is indeterminate, then the limit of the quotient of their derivatives will have the same value.

When is L'Hospital's Rule applicable?

L'Hospital's Rule is applicable when the limit of a function can be written as a quotient of two functions, both of which approach 0 or ∞ as the independent variable approaches a certain value. It is also applicable when the limit of a function can be written as a quotient of two functions, one of which approaches 0 and the other approaches ∞ as the independent variable approaches a certain value.

How do you use L'Hospital's Rule to solve a limit?

To use L'Hospital's Rule, first check if the limit is indeterminate. If it is, take the derivative of the numerator and denominator separately. Then, evaluate the limit of the derivative quotient. If the limit is still indeterminate, repeat the process until the limit can be evaluated. Finally, the limit of the original function will have the same value as the limit of the derivative quotient.

Can L'Hospital's Rule be used for limits approaching finite values?

No, L'Hospital's Rule can only be used for limits approaching ∞ or 0. If a limit is approaching a finite value, other methods such as algebraic manipulation or substitution should be used to evaluate the limit.

Are there any limitations to using L'Hospital's Rule?

Yes, L'Hospital's Rule may not work for all limits and it is important to check the conditions of applicability before using it. Additionally, it may not always provide the correct answer, so it is important to double check the result using other methods.

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