Limit of arctan(C tan(x))

**JulieK** · June 6th 2010, 02:24 PM

Hello

What is

$\mathrm{\lim_{x\rightarrow\frac{\pi}{2}}}\arctan\l eft[C\tan(x)\right]$

where $C$ is a constant. My guess is that it is $-\frac{\pi}{2}$ for $C < 0$ and $\frac{\pi}{2}$ for $C > 0$ . However, if this is correct I require more technical treatment.

Many thanks in advance.

**JulieK** · June 7th 2010, 01:46 AM

I evaluated this limit numerically and it obviously converges to $\pi/2$ when $C > 0$ and to $- \pi/2$ when $C < 0$ as $x$ converges to $\pi / 2$ . However, I am still looking for a formal proof.
Thanks.

**Sudharaka** · June 7th 2010, 02:50 AM

Originally Posted by JulieK

Hello

What is

$\mathrm{\lim_{x\rightarrow\frac{\pi}{2}}}\arctan\l eft[C\tan(x)\right]$

where $C$ is a constant. My guess is that it is $-\frac{\pi}{2}$ for $C < 0$ and $\frac{\pi}{2}$ for $C > 0$ . However, if this is correct I require more technical treatment.

Many thanks in advance.

Dear JulieK,

The limit does not exist.

Notice that, $\lim_{x\rightarrow{\frac{\pi}{2}}^+}(\tan{x})\neq{ \lim_{x\rightarrow{\frac{\pi}{2}}^{-}}(\tan{x})}$

**JulieK** · June 7th 2010, 03:08 AM

Dear Sudharaka

Thank you for your reply. Yes, you are right. But what about one-sided limits. In fact I am mainly interested in $\lim_{x\rightarrow{\frac{\pi}{2}}^-}(\arctan(C\tan{x}))$ and that is what I numerically evaluated.

**Sudharaka** · June 7th 2010, 05:36 PM

Originally Posted by JulieK

Dear Sudharaka

Thank you for your reply. Yes, you are right. But what about one-sided limits. In fact I am mainly interested in $\lim_{x\rightarrow{\frac{\pi}{2}}^-}(\arctan(C\tan{x}))$ and that is what I numerically evaluated.

Dear JulieK,

$\lim_{x\rightarrow{\frac{\pi}{2}}^-}tan^{-1}(C\tan{x})=tan^{-1}\left[c\lim_{x\rightarrow{\frac{\pi}{2}}^-}\tan{x}\right]$ ; Since $tan^{-1}$ is continuous.

$If~c>0\Rightarrow~y=c\lim_{x\rightarrow{\frac{\pi} {2}}^-}\tan{x}=\infty$

Therefore, $\lim_{x\rightarrow{\frac{\pi}{2}}^-}tan^{-1}(C\tan{x})=\lim_{y\rightarrow\infty}tan^{-1}y=\frac{\pi}{2}$

$If~c<0\Rightarrow~y=c\lim_{x\rightarrow{\frac{\pi} {2}}^-}\tan{x}=-\infty$

Therefore, $\lim_{x\rightarrow{\frac{\pi}{2}}^-}tan^{-1}(C\tan{x})=\lim_{y\rightarrow~-\infty}tan^{-1}y=-\frac{\pi}{2}$

Hope this will help you.

**JulieK** · June 8th 2010, 12:41 AM

Dear Sudharaka
Thank you very much.
Best wishes
JulieK

**Sudharaka** · June 8th 2010, 12:54 AM

Originally Posted by JulieK

Dear Sudharaka
Thank you very much.
Best wishes
JulieK

Dear JulieK,

You are welcome.

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