3) We have
[tex]f(x) = \sec\left(\dfrac{\pi x}2\right) = \dfrac1{\cos\left(\frac{\pi x}2\right)}[/tex]
which has vertical asymptotes (i.e. infinite discontinuities) whenever the denominator is zero. This happens for
[tex]\cos\left(\dfrac{\pi x}2\right) = 0[/tex]
[tex]\implies \dfrac{\pi x}2 = \cos^{-1}(0) + 2n\pi \text{ or } \dfrac{\pi x}2 = -\cos^{-1}(0) + 2n\pi[/tex]
(where [tex]n[/tex] is any integer)
[tex]\implies \dfrac{\pi x}2 = \dfrac\pi2 + 2n\pi \text{ or } \dfrac{\pi x}2 = -\dfrac\pi2 + 2n\pi[/tex]
[tex]\implies x = 1 + 4n \text{ or } x = -1 + 4n[/tex]
So the graph of [tex]f(x)[/tex] has vertical asymptotes whenever [tex]x=4n\pm1[/tex] and [tex]n\in\Bbb Z[/tex].
4) Given
[tex]h(t) = \begin{cases} t^3+1 & \text{if } t<1 \\ \frac12 (t+1) & \text{if } t\ge1 \end{cases}[/tex]
we have the one-sided limits
[tex]\displaystyle \lim_{t\to1^-} h(t) = \lim_{t\to1} (t^3+1) = 1^3+1 = 2[/tex]
and
[tex]\displaystyle \lim_{t\to1^+} h(t) = \lim_{h\to1} \frac{t+1}2 = \frac{1+1}2 = 1[/tex]
The one-sided limits don't match, so the two-sided limit [tex]L[/tex] does not exist. In other words, the limit does not exist at [tex]x=1[/tex] because the function approaches different values from the left and right side of [tex]x=1[/tex].
