L'hopital's rule, indeterminate forms

[Panphobia]

Homework Statement

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

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.

 

[MisterX]

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.

 

[scurty]

Homework Statement

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

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.