Is $(\frac{1}{x})\cos(\frac{1}{x})$ continuous? $$f(x) = \begin{cases}\frac{1}{x} \,\cos\frac {1}{x} &\text{if } x>0 \\
0 & \text{if } x = 0 \end{cases}$$
Is this function continuous?
My intuition says no because as $x$ approaches $0$, $f(x)$ approaches $\infty$. 
Is that a good enough reason?
What about $f(x) = \left(\frac{1}{x}\right)^a$ where $a $ is positive ?
Added: I'm sorry I was trying to mean $f(x) = \frac{1}{x}\,\cos \big(\frac{1}{x}\big)^c$ where $c $ is positive. Would you please help me with that ?

Yes, it is good enough reason. To formalize it just notice that for $f$ to be continuous at $0$ we would need $$0=f(0)=\lim_{k\to+\infty}f\left(\frac{1}{2k\pi}\right)=\lim_{k\to+\infty}2k\pi=+\infty,$$
which is absurd.
For your second question, $f(x)=(1/x)^a=e^{-a\log x}$ is continuous in $(0,+\infty)$ but not at $0$ because $f$ is not bounded near $x=0.$
For your modified second question: $f(x)=\frac{1}{x}\cos\left(\frac{1}{x}\right)^c$, $c>0$, is also not continuous at $x=0$ because the same reasoning as above applies and leads to the contradiction $0=+\infty.$
