arctan and limit problem

[Stereotypical]

Arctangent,infinity and limit problem

Greetings, this is my first thread & post.
I have a problem:

I want to find the limit of tan^-1(e^x) as x approaches infinity.
Specifically:
lim_x->+∞(tan^-1e^x)

I already know that it's π/2, but why? How to prove it scientifically?

Can I use squeeze/sandwich theorem? If so, how? I don't know how to use it with inverse trigonometric functions such as tan^-1x .

Thanks in advance.
Happy new year 2012, too.

 

[galactus]

Note that $\displaystyle \lim_{x\to \infty}e^{x}=\infty$

So, we have $\displaystyle tan^{-1}(\infty)=\frac{\pi}{2}$

 

[Stereotypical]

Note that $\displaystyle \lim_{x\to \infty}e^{x}=\infty$

So, we have $\displaystyle tan^{-1}(\infty)=\frac{\pi}{2}$

Hi and thank you for your response.

I already knew that $\displaystyle \lim_{x\to \infty}e^{x}=\infty$ .

The question was: how to prove that $\displaystyle tan^{-1}(\infty)=\frac{\pi}{2}$ ? With squeeze theorem??

 

[galactus]

arctan has an upper bound of $\displaystyle \frac{\pi}{2}$.

I doubt if you need the squeeze theorem.

 

[pka]

The question was: how to prove that $\displaystyle tan^{-1}(\infty)=\frac{\pi}{2}$ ? With squeeze theorem??

I agree with reply #4: the squeeze theorem does not really apply here.
We know that $\displaystyle \arctan (x):\mathbb{R} \to \left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right)$ is a bijection (one-to-one & onto).
Therefore, $\displaystyle \arctan$ is increasing function so \(\displaystyle \[\mathop {\lim }\limits_{x \to \infty } \arctan(x)=\frac{\pi}{2}.\)