How do I evaluate $\lim_{x\to \infty} tan^{-1}(e^x)$ How do I evaluate the following limit ? What things do I need to know to solve limit like the following ?$$\lim_{x\to \infty} \tan^{-1}(e^x)$$
 A: 
Recall the identity 
$$\arctan(t)+\arctan(1/t)=\pi/2 \tag 1$$ 
for $t>0$.  

Therefore, substituting $t=e^x$ in $(1)$, rearranging, and taking the limit as $x\to \infty$, we obtain
$$\begin{align}\lim_{x\to \infty}\arctan(e^x)&=\lim_{x\to \infty}\left(\pi/2-\arctan(e^{-x})\right)\\\\
&=\pi/2-\lim_{x\to \infty}\arctan(e^{-x})\\\\
&=\pi/2-\arctan(\lim_{x\to \infty}e^{-x})\\\\
&=\pi/2-\arctan(0)\\\\
&=\pi/2
\end{align}$$
A: Note that, as $x\to\infty$, $e^x\to\infty$, so we reduce this to
$$\lim_{y\to\infty} \tan^{-1}(y)$$
So, we want to find a value of $\theta$ such that
$$\tan(\theta) = \infty$$
(rigorously, we require that $\lim_{\phi\to\theta} \tan(\phi) = \infty$). Since
$$\tan(\theta) = \frac{\sin\theta}{\cos\theta}$$
we need $\sin\theta>0$, $\cos\theta=0$. This happens at 
$$\theta = \frac{\pi}{2}$$
so that is the value of our limit. 
A: As $x \to \infty$ then $e^x \to \infty$:
Therefore by the substitution $t=e^x$ our limit becomes:
$$=\lim_{t \to \infty} \arctan (t)$$
Note that by definition $\arctan \in [\frac{\pi}{2},\frac{-\pi}{2}]$. Now imagine the angle $\theta$ counterclockwise from the positive $x$ axis made by a line going through the origin with slope $m$ that lies in the 1st quadrant.  You can think of the line line has a rise of $m$ from some point to another and a run of $1$ to that same point to another. The run is the adjacent, the rise is the opposite side. Therefore we have $\tan (\theta)=m$. As we are in the first quadrant, we have $\theta=\arctan(m)$. As the slope of the line increases we have that the slope approaches becoming vertical, i.e. $\theta$ approaches an angle of $\frac{\pi}{2}$.

As a test of understanding :
Now think about a line in the 4th quadrant with slope $m$, that passes through the origin creating some counterclockwise angle $\theta$ with the positive $x$ axis. What will be happen to $\theta$ as $m \to -\infty$.


Do you have an intuition for what is happening? 
The angle is approaching $-\frac{\pi}{2}$ hence $\lim_{t \to -\infty} \arctan (t)=-\frac{\pi}{2}$.
Going back to the original question Another way to look at it is:
As,
$$\lim_{t \to \frac{\pi}{2}^-} \tan (t)=\lim_{t \to \frac{\pi}{2}^-} \frac{\sin (t)}{\cos (t)}=\infty$$
We have:
$$\lim_{x \to \infty} \arctan (x)=\frac{\pi}{2}$$
A: This is not an answer but it is too long for a comment.
Since Dr. MV already gave the trick, that is to say that, for any positive value of $x$, $$\tan^{-1}(e^x)=\frac \pi 2-\tan^{-1}(e^{-x})$$ let me suppose that you want to compute the numerical value.
Using $$\tan^{-1}(t)=t-\frac{t^3}{3}+\frac{t^5}{5}+O\left(t^7\right)$$ then $$\tan^{-1}(e^x)=\frac \pi 2-e^{-x}+\frac 13 e^{-3x}-\frac 15 e^{-5x}+\cdots$$
Let me try using $x=10$; the approximation would then give $$\tan^{-1}(e^{10})=\frac \pi 2-e^{-10}+\frac 13 e^{-30}-\frac 15 e^{-50}\approx\color{red}{1.570750926865165326456310325709}2480$$ while the exact value would be (for $35$ significant figures) $$\color{red}{1.5707509268651653264563103257093048}$$
This is the method I would use if I had to write a program.
