Simplifying logs

**janedoe** · September 12th 2010, 02:42 PM

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

Thanks!

**Plato** · September 12th 2010, 02:57 PM

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

**janedoe** · September 12th 2010, 04:08 PM

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

**Plato** · September 12th 2010, 04:15 PM

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.

**Soroban** · September 12th 2010, 04:52 PM

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

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