Part (a)
We need to consider
[tex]x=\frac{1}{k\pi}, k \in \mathbb{Z}[/tex]
[tex]f\left(\frac{1}{k\pi} \right)=\tan (k\pi)
[/tex]
tan(x) has a period of pi, so this is equal to tan(0)=0.
Part (b)
We need to consider
[tex]x=\frac{4}{(4n+1)\pi}, n \in \mathbb{Z} \\ \\ f \left(\frac{4}{(4n+1)\pi} \right)=\tan \left(\frac{(4n+1)\pi}{4} \right) \\ \\ =\tan \left(\pi n+\frac{\pi}{4} \right) \\ \\ =1[/tex]
Part (c)
We see that 1/x approaches infinity, and hence the limit does not exist.
