1. ## Simplifying logs

see this pic: http://i56.tinypic.com/2j5fgwx.jpg

Thanks!

2. The limit of logs is log of limits.
$\displaystyle \lim _{x \to a} \ln (f(x)) = \ln (\lim _{x \to a} f(x))$

3. but mine is different....there is simply no ln in step 2...

4. Originally Posted by janedoe
but mine is different....there is simply no ln in step 2...
But the limit in step 2 determines the limit in step one.

5. Hello, janedoe!

I'm sure it was a typo . . .

It should have said:

$\displaystyle \lim_{x\to\infty}\bigg[x - \ln(x^2+1)\bigg] \;=\;\lim_{x\to\infty}\bigg[\ln(e^x) - \ln(x^2+1)\bigg]$

. . $\displaystyle =\;\lim_{x\to\infty}\bigg[\ln\left(\frac{e^x}{x^2+1}\right)\bigg] \;=\;\ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{x^2+1}\right)\bi gg]$

Apply L'Hopital: . $\displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2x}\right)\bigg]$

Apply L'Hopital: . $\displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2}\right)\bigg] \;=\;\ln(\infty) \;=\;\infty$

Therefore: . $\displaystyle \lim_{x\to\infty}\bigg[x - \ln(x^2+1)\bigg] \;=\;\infty$