The limit of an inverse trig function. $$\lim_{x\to\infty}\tan^{-1}(\dfrac{x}{4})$$
I have a question about why this limit is $\pi/2$. If the argument of $\tan^{-1}$ goes to infinity, doesn't the slope (since it is the tangent function) also go to infinity? I don't see why it would go to $\pi/2$. Could someone explain to me why this is? Also what if instead of $\tan$, we had $\sin$?
 A: I think your main problem is with the notion:
$$ \lim\limits_{x \to \infty} \tan^{-1} (\frac{x}{4}) \neq \lim\limits_{x \to \infty} \frac{1}{\tan (\frac{x}{4})}$$
instead if $f(x) = \tan (x)$
$$\tan^{-1} (x) = f^{-1}(x)$$
For clarity I will use $\tan^{-1} (x) = \arctan (x)$
Knowing that let's take the limit:
First lets substitute $t = \frac{x}{4}$ (as suggested before):
$$ \lim\limits_{x \to \infty} \arctan (\frac{x}{4}) = \lim\limits_{t \to \infty} \arctan (t) , t = \frac{x}{4}$$
notice that $$ \lim\limits_{t \to \infty} t = \lim\limits_{x \to \infty} \frac{x}{4} = \infty $$
now we notice that since the $\arctan (x) $ function should produce a number that if input into the tangent function will output $x$, and since the tangent function has a range from $(-\infty , \infty)$ between $ (\frac{-\pi}{2},\frac{\pi}{2}) $, the inverse tangent if given an input from $(-\infty , \infty)$ you will get a value from  $ (\frac{-\pi}{2},\frac{\pi}{2}) $. As we take $ t \to \infty$, we get closer and closer to $\frac{\pi}{2}$, therefore:
$$ \lim\limits_{x \to \infty} \arctan (\frac{x}{4}) = \lim\limits_{t \to \infty} \arctan(t)  = \frac{\pi}{2} $$
A: Hint: Make the substitution $t=\frac x4,$ noting that $t\to\infty$ precisely as $x\to\infty,$ so that we can rewrite as $$\lim_{t\to\infty}\tan^{-1}(t).$$ Now, pay close attention to how the inverse tangent function is defined. In particular, it is the inverse of the restriction of the tangent function to which interval?
