# Find the limit - inverse trig

• Jan 12th 2010, 11:10 AM
DBA
Find 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 AM
Calculus26
lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2
• Jan 12th 2010, 11:54 AM
Drexel28
Quote:

Originally Posted by DBA
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

I sure will give you another limit equation, son. If we let $\ln(x)=z$ this becomes $\lim_{z\to-\infty}\text{arctan}(z)$. Is the answer clear now?
• Jan 12th 2010, 12:23 PM
galactus
Let's do something wacky and fun(Wink).

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

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

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

Use L'hopital on the inside:

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

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

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

So, we get $\frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}$
• Jan 12th 2010, 12:54 PM
DBA
Quote:

Originally Posted by Calculus26
lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[lim(ln(x)] = tan^(-1)(-infinity)= -pi/2

is lim x-> 0+ tan ^-1 (lnx) = tan^(-1)[limx-> 0+ (ln(x))] ?
• Jan 12th 2010, 12:56 PM
DBA
Quote:

Originally Posted by Drexel28
I sure will give you another limit equation, son. If we let $\ln(x)=z$ this becomes $\lim_{z\to-\infty}\text{arctan}(z)$. Is the answer clear now?

Hi, thanks.
but why lim z -> -infinity?
• Jan 12th 2010, 12:58 PM
DBA
Quote:

Originally Posted by galactus
Let's do something wacky and fun(Wink).

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

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

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

Use L'hopital on the inside:

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

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

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

So, we get $\frac{i}{2}\cdot {\pi}i=\frac{\pi}{2}(-1)=\frac{-\pi}{2}$

Thanks, but we did not have L'hopital yet and with the number i I don't understand a thing. Sorry
• Jan 12th 2010, 01:00 PM
Calculus26
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 PM
galactus
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 $tan^{-1}({\infty})=\frac{\pi}{2}$