- #1
redsox5
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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.
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.
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.
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]
redsox5 said:wait..i the correct answer -1 since it equals infin/neg infin
redsox5 said:how many time do you have to use l'hospital's rule..it's getting really messy
redsox5 said:but it's 1/0 can I still use l'hospitals'?
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.
redsox5 said:your numerator is wrong, her's the oringal again
lim x-> infin x[(x+1)ln(1+(1/x))-1]
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.
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.
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.
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.
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.