Question:
Find $\lim _{x \rightarrow 1} f(x)$, where $f(x)= \begin{cases}x^{2}-1, & x \leq 1 \\ -x^{2}-1, & x>1\end{cases}$
Solution:
The given function is
$f(x)=\left\{\begin{array}{l}x^{2}-1, x \leq 1 \\ -x^{2}-1, x>1\end{array}\right.$
$\lim _{1} f(x)=\lim \left[x^{2}-1\right]=1^{2}-1=1-1=0$
$\lim _{x \rightarrow 1^{\prime}} f(x)=\lim _{x \rightarrow 1}\left[-x^{2}-1\right]=-1^{2}-1=-1-1=-2$
It is observed that $\lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$.
Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.
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