Question

the graph of the function ( f(x)=\tan x ) is given above for the interval ( x in0,2 pi ) only. determine the one - sided limit. then indicate the equation of the vertical asymptote. find ( lim _{x
ightarrowleft(\frac{pi}{2}
ight)^{+}} f(x)= ) this indicates the equation of a vertical asymptote is ( x = ) find ( lim _{x
ightarrowleft(\frac{3 pi}{2}
ight)^{-}} f(x)= ) this indicates the equation of a vertical asymptote is ( x = ) question help: video message instructor submit question jump to answer

Explanation:

Step1: Analyze the limit as \(x\to(\frac{\pi}{2})^+\)

As \(x\) approaches \(\frac{\pi}{2}\) from the right (\(x\to(\frac{\pi}{2})^+\)), \(\tan x=\frac{\sin x}{\cos x}\). When \(x\to(\frac{\pi}{2})^+\), \(\cos x\to0^-\) and \(\sin x = 1\). So \(\lim_{x\to(\frac{\pi}{2})^+}\tan x=-\infty\). The vertical - asymptote occurs where the function is undefined (denominator of \(\tan x=\frac{\sin x}{\cos x}\) is \(0\)). When \(\cos x = 0\) and \(x\in[0,2\pi]\), \(x = \frac{\pi}{2}\) is a vertical asymptote.

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

As \(x\to(\frac{3\pi}{2})^-\), \(\cos x\to0^+\) and \(\sin x=- 1\). So \(\lim_{x\to(\frac{3\pi}{2})^-}\tan x=\frac{\sin x}{\cos x}=-\infty\). When \(\cos x = 0\) and \(x\in[0,2\pi]\), \(x=\frac{3\pi}{2}\) is a vertical asymptote.

Answer:

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