Question

I need Help with calculus please, thank you! 10points

Answer 1

3) We have

$f(x) = \sec\left(\dfrac{\pi x}2\right) = \dfrac1{\cos\left(\frac{\pi x}2\right)}$

which has vertical asymptotes (i.e. infinite discontinuities) whenever the denominator is zero. This happens for

$\cos\left(\dfrac{\pi x}2\right) = 0$

$\implies \dfrac{\pi x}2 = \cos^{-1}(0) + 2n\pi \text{ or } \dfrac{\pi x}2 = -\cos^{-1}(0) + 2n\pi$

(where $n$ is any integer)

$\implies \dfrac{\pi x}2 = \dfrac\pi2 + 2n\pi \text{ or } \dfrac{\pi x}2 = -\dfrac\pi2 + 2n\pi$

$\implies x = 1 + 4n \text{ or } x = -1 + 4n$

So the graph of $f(x)$ has vertical asymptotes whenever $x=4n\pm1$ and $n\in\Bbb Z$.

4) Given

$h(t) = \begin{cases} t^3+1 & \text{if } t$

we have the one-sided limits

$\displaystyle \lim_{t\to1^-} h(t) = \lim_{t\to1} (t^3+1) = 1^3+1 = 2$

and

$\displaystyle \lim_{t\to1^+} h(t) = \lim_{h\to1} \frac{t+1}2 = \frac{1+1}2 = 1$

The one-sided limits don't match, so the two-sided limit $L$ does not exist. In other words, the limit does not exist at $x=1$ because the function approaches different values from the left and right side of $x=1$.