Question

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

Answer 1

3) We have

`f(x) = \sec\left(\frac{\pi x}2\right) = \frac1{\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(\frac{\pi x}2\right) = 0`

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

(where
`n` is any integer)

`\implies \frac{\pi x}2 = \frac\pi2 + 2n\pi \text{ or } \frac{\pi x}2 = -\frac\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<1 \\ \frac12 (t+1) & \text{if } t\ge1 \end{cases}`

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`.