see this pic: http://i56.tinypic.com/2j5fgwx.jpg
Thanks!
see this pic: http://i56.tinypic.com/2j5fgwx.jpg
Thanks!
Hello, janedoe!
I'm sure it was a typo . . .
It should have said:
$\displaystyle \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] $
. . $\displaystyle \displaystyle =\;\lim_{x\to\infty}\bigg[\ln\left(\frac{e^x}{x^2+1}\right)\bigg] \;=\;\ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{x^2+1}\right)\bi gg]$
Apply L'Hopital: .$\displaystyle \displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2x}\right)\bigg] $
Apply L'Hopital: .$\displaystyle \displaystyle \ln\bigg[\lim_{x\to\infty}\left(\frac{e^x}{2}\right)\bigg] \;=\;\ln(\infty) \;=\;\infty $
Therefore: .$\displaystyle \displaystyle \lim_{x\to\infty}\bigg[x - \ln(x^2+1)\bigg] \;=\;\infty$