the graph of the function $f(x)=\tan x$ is given above for the interval $xin0,2pi$ only. determine the one…

the graph of the function $f(x)=\tan x$ is given above for the interval $xin0,2pi$ only. determine the one - sided limit. then indicate the equation of the vertical asymptote. find $lim_{x\rightarrow(\frac{pi}{2})^{-}}f(x)=$\nthis indicates the equation of a vertical asymptote is $x =$\nfind $lim_{x\rightarrow(\frac{3pi}{2})^{-}}f(x)=$\nthis indicates the equation of a vertical asymptote is $x =$\nquestion help: video message instructor

the graph of the function $f(x)=\tan x$ is given above for the interval $xin0,2pi$ only. determine the one - sided limit. then indicate the equation of the vertical asymptote. find $lim_{x\rightarrow(\frac{pi}{2})^{-}}f(x)=$\nthis indicates the equation of a vertical asymptote is $x =$\nfind $lim_{x\rightarrow(\frac{3pi}{2})^{-}}f(x)=$\nthis indicates the equation of a vertical asymptote is $x =$\nquestion help: video message instructor

Answer

Explanation:

Step1: Recall tangent - function property

The tangent function (y = \tan x=\frac{\sin x}{\cos x}). As (x\to\frac{\pi}{2}^{-}), (\sin x\to1) and (\cos x\to0^{+}). Since (\tan x=\frac{\sin x}{\cos x}), (\lim_{x\to(\frac{\pi}{2})^{-}}\tan x=\infty). The vertical - asymptote occurs when (\cos x = 0), and for (x\to\frac{\pi}{2}^{-}), the equation of the vertical asymptote is (x = \frac{\pi}{2}).

Step2: Analyze (x\to\frac{3\pi}{2}^{-})

As (x\to\frac{3\pi}{2}^{-}), (\sin x\to - 1) and (\cos x\to0^{+}). Since (\tan x=\frac{\sin x}{\cos x}), (\lim_{x\to(\frac{3\pi}{2})^{-}}\tan x=-\infty). The vertical - asymptote occurs when (\cos x = 0), and for (x\to\frac{3\pi}{2}^{-}), the equation of the vertical asymptote is (x=\frac{3\pi}{2}).

Answer:

(\lim_{x\to(\frac{\pi}{2})^{-}}f(x)=\infty), (x = \frac{\pi}{2}), (\lim_{x\to(\frac{3\pi}{2})^{-}}f(x)=-\infty), (x=\frac{3\pi}{2})