Browse Questions

# Find $\lim\limits_{ x \to 1} f(x)$, where $f(x)=\left\{\begin{array}{1 1}x^2-1, &x \leq 1\\-x^2-1, &x>1\end{array}\right.$

The given function is
$f(x)=\left\{\begin{array}{1 1}x^2-1, &x \leq 1\\-x^2-1, &x>1\end{array}\right.$
$\lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1} [ x^2-1] = 1^2-1 = 1-1 = 0$
$\lim\limits_{x \to 1^+} f(x) = \lim\limits_{x \to 1} [ -x^2-1] = -1^2-1 = -1-1 = -2$
It is observed that $\lim\limits_{x \to 1^-}f(x) \neq \lim\limits_{x \to 1^+}f(x).$
Hence, $\lim\limits_{x \to 1}f(x)$ does not exist.