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}.\)