find the limit. if in doubt, look at the functions graph.\n\\( \\lim _{x \\rightarrow \\infty} \\tan ^{-1} x…

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

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

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})