Answer
No, the function is not differentiable at x = 1.
Work Step by Step
Derivative from the left:
$\lim\limits_{x \to 1^{-}} f(x)$
$= \lim\limits_{x \to 1^{-}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 1^{-}} \frac{x - 1}{x - 1}$
$= \lim\limits_{x \to 1^{-}} 1$
$= 1$
Derivative from the right:
$\lim\limits_{x \to 1^{+}} f(x)$
$=\lim\limits_{x \to 1^{+}} \frac{f(x) - f(c)}{x - c}$
$= \lim\limits_{x \to 1^{+}} \frac{x^{2} - 1}{x - 1}$
$= \lim\limits_{x \to 1^{+}} \frac{(x + 1)(x - 1)}{x - 1}$
$= \lim\limits_{x \to 1^{+}} (x + 1)$
$= 2$
Since $\lim\limits_{x \to 1^{-}} f(x) \ne \lim\limits_{x \to 1^{+}} f(x)$, the function $f(x)$ is not differentiable at x = 1.
