Answer
a) $\dfrac{x^2-1}{x-1}$.
b) $\dfrac{x^2-1}{x-1}$.
c) $\dfrac{x+1}{x-1}$.
Work Step by Step
a) Let $f(x)=\dfrac{x^2-1}{x-1}$.
For $c=1$, $f(c)=\dfrac{1^2-1}{1-1}=\dfrac{0}{0}$, fo $f(1)$ is indeterminate, but the left and right limits are equal (to $2$).
b) Consider the same function $f(x)=\dfrac{x^2-1}{x-1}$.
For $c=1$ the left and right limits when $x\rightarrow 1$ are equal (to $2$), so the limit of $f$ when $x\rightarrow 1$ exists, but $f(1)$ is indeterminate.
c) Let $f(x)=\dfrac{x+1}{x-1}$.
For $c=1$, $f$ is undefined, but $f(1)=\dfrac{1+1}{1-1}=\dfrac{2}{0}$ is not in indeterminate form.
