Question

find the limit. if in doubt, look at the functions graph.
\\( \lim _{x \
ightarrow \infty} \tan ^{-1} x \\)
\\( \lim _{x \
ightarrow \infty} \tan ^{-1} x= \\) (type an exact answer.)

Explanation:

Step1: Recall the range and behavior of \(y = \tan^{-1}x\)

The function \(y=\tan^{-1}x\) has a range of \((-\frac{\pi}{2},\frac{\pi}{2})\). As \(x\) approaches \(+\infty\), we consider the behavior of the inverse - tangent function.
We know that \(\tan\theta=x\), and when \(x\to+\infty\), if \(\theta = \tan^{-1}x\), then \(\lim_{x\to+\infty}\tan^{-1}x\) corresponds to the value of \(\theta\) for which \(\tan\theta\to+\infty\) and \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\).

Step2: Determine the limit value

Since \(\tan(\frac{\pi}{2}-\epsilon)\to+\infty\) as \(\epsilon\to0^{+}\) and \(\frac{\pi}{2}-\epsilon\in(-\frac{\pi}{2},\frac{\pi}{2})\) for small positive \(\epsilon\), we have \(\lim_{x\to+\infty}\tan^{-1}x=\frac{\pi}{2}\)

Answer:

\(\frac{\pi}{2}\)