indeterminate Forms Prob: lim[lnx - ln(1+x)] x->infty

[hank]

I'm stuck on this one, and I'm not sure where to go at all...

Find the limit of:

lim x->infinity [lnx - ln(1 + x)]

I'm sure this is an indeterminate form of infinity - infinity, but that's as far as I can get.

I know I need to convert this to a form of infinity / infinity.

Which leads me to:

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

This is as far as I can get tho, and I'm not sure this is right.

Can someone point me in the right direction?

 

[skeeter]

you're almost there ...

lim{x->infinity} ln[x/(1 + x)]

as x->infinity, what value does x/(1+x) approach?

what is the natural log of that value?

 

[soroban]

Re: indeterminate Forms Problem

Hello, hank!

\(\displaystyle \lim_{x\to\infty}\left[\ln x\, -\, \ln(1 + x)\right]\)

I know I need to convert this to a form of \(\displaystyle \frac{\infty}{\infty}\)

which leads me to: \(\displaystyle \:\lim_{x\to\infty}\ln\left(\frac{x}{1\,+\,x}\right)\)

In the fraction, divide numerator and denominator by \(\displaystyle x\)

.\(\displaystyle \L\lim_{x\to\infty}\ln\left(\frac{\frac{x}{x}}{\frac{1}{x}\,+\,\frac{x}{x}}\right) \:=\:\lim_{x\to\infty}\ln\left(\frac{1}{\frac{1}{x}\,+\,1}\right) \:=\:\ln\left(\frac{1}{0\,+\,1}\right)\:=\:\ln(1)\:=\:0\)

 

[hank]

Thank you so much!

The thing I love about math is that when you get stuck somewhere someone else can come along and see the thing you didn't. Then when you finally see it, you're like DUH! I *knew* that!

Thanks everyone!