Find the limit:

lim x-> 0+ tan ^-1 (lnx) =?

I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...

Thanks

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- Jan 12th 2010, 11:10 AMDBAFind the limit - inverse trig
Find the limit:

lim x-> 0+ tan ^-1 (lnx) =?

I have no ide where to begin here? Can someone explain this in simple steps? Please, do not just give me another limit equation...

Thanks - Jan 12th 2010, 11:54 AMCalculus26
lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2

- Jan 12th 2010, 11:54 AMDrexel28
- Jan 12th 2010, 12:23 PMgalactus
Let's do something wacky and fun(Wink).

Note that $\displaystyle tan^{-1}(u)=\frac{i}{2}\left(ln(1-ix)-ln(1+ix)\right)$

$\displaystyle =\frac{i}{2}ln(\frac{1-ix}{1+ix})$

$\displaystyle =\frac{i}{2}ln\left(\lim_{x\to 0}\frac{2-2i\cdot ln(x)}{1+(ln(x))^{2}}-1\right)$

Use L'hopital on the inside:

$\displaystyle \frac{i}{2}ln\left(\lim_{x\to 0}\frac{-i}{ln(x)}-1\right)$

The limit tends to -1 and we get $\displaystyle \frac{i}{2}ln(-1)$

$\displaystyle ln(-1)={\pi}i$

So, we get $\displaystyle \frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}$ - Jan 12th 2010, 12:54 PMDBA
- Jan 12th 2010, 12:56 PMDBA
- Jan 12th 2010, 12:58 PMDBA
- Jan 12th 2010, 01:00 PMCalculus26
yes in general

limf(g) = f(lim(g)) at a point if x = a if g is continuous at x = a and the limf exists at g(a) - Jan 12th 2010, 01:01 PMgalactus
I am sorry. Perhaps I shouldn't have done that. Just having fun. You will learn L'Hopital in time. It's a handy tool.

There are easier ways to go about it. Take note that $\displaystyle tan^{-1}({\infty})=\frac{\pi}{2}$