L'hopital's rule, indeterminate forms

  • Thread starter Panphobia
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  • #1
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Homework Statement



[itex]lim_{x -> \infty} \left( \frac{x}{x+1} \right) ^ {x}[/itex]



The Attempt at a Solution



So I did e^whole statement with ln(x/(x+1))*x, after that I multiplied that expression by 1/x/1/x, then I go ln(x/(x+1)/1/x, I tried taking derivative of top and bottom but it doesn't help with finding an answer.
 
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Answers and Replies

  • #2
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So I did e^whole statement with ln(x/(x+1))*x, after that I multiplied that expression by 1/x/1/x, then I go ln(x/(x+1)/1/x, I tried taking derivative of top and bottom but it doesn't help with finding an answer.

It did for me. Try it again, maybe.
 
  • #3
392
17

Homework Statement



[itex]lim_{x -> \infty} \left( \frac{x}{x+1} \right) ^ {x}[/itex]



The Attempt at a Solution



So I did e^whole statement with ln(x/(x+1))*x, after that I multiplied that expression by 1/x/1/x, then I go ln(x/(x+1)/1/x, I tried taking derivative of top and bottom but it doesn't help with finding an answer.

I wouldn't e^ln(statement).I would write it as

##y= \displaystyle \lim_{x \to \infty} \left( \frac{x}{x+1} \right)^x##
##\ln{y}= \displaystyle \lim_{x \to \infty} \ln{\left[\left( \frac{x}{x+1} \right)^x\right]}##

Then you eventually solve for y. What do you get when you perform l'Hospital's rule? This is the correct procedure so far.
 

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