Let $f(x)=\frac{1}{x-3}$ on $(3,4]$. I need to show that f is continuous in this interval, but not uniformly continuous.
Idea: Since $f'(x)=\frac{-1}{(x-3)^{2}}$, the derivative exists at all points in $(3,4]$ then $f$ is continuous in $(3,4]$ But if we look at uniform continuity, $\lim_{x \to a^{+}}f(x)$ is infinity while $f(3)$ is simply undefined. Is this accurate to say? More worried about the second part.