Question

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

Explanation:

Step1: Recall the definition of secant

Recall that $\sec x=\frac{1}{\cos x}$.

Step2: Analyze $\lim_{x

ightarrow(\frac{\pi}{2})^{-}}\sec x$
As $x
ightarrow(\frac{\pi}{2})^{-}$, $\cos x
ightarrow0^{+}$ (since $\cos x$ is positive and approaching 0 as $x$ approaches $\frac{\pi}{2}$ from the left). Then $\lim_{x
ightarrow(\frac{\pi}{2})^{-}}\sec x=\lim_{x
ightarrow(\frac{\pi}{2})^{-}}\frac{1}{\cos x}=+\infty$. The vertical - asymptote occurs when $\cos x = 0$, and for $x
ightarrow\frac{\pi}{2}$, the equation of the vertical asymptote is $x = \frac{\pi}{2}$.

Step3: Analyze $\lim_{x

ightarrow(\frac{3\pi}{2})^{-}}\sec x$
As $x
ightarrow(\frac{3\pi}{2})^{-}$, $\cos x
ightarrow0^{-}$ (since $\cos x$ is negative and approaching 0 as $x$ approaches $\frac{3\pi}{2}$ from the left). Then $\lim_{x
ightarrow(\frac{3\pi}{2})^{-}}\sec x=\lim_{x
ightarrow(\frac{3\pi}{2})^{-}}\frac{1}{\cos x}=-\infty$. The vertical - asymptote occurs when $\cos x = 0$, and for $x
ightarrow\frac{3\pi}{2}$, the equation of the vertical asymptote is $x=\frac{3\pi}{2}$.

Answer:

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