Question
Find $\lim _{x \rightarrow 1} f(x)$, where $f(x)=\left\{\begin{array}{l}x^{2}-1, x \leq 1 \\ -x^{2}-1, x>1\end{array}\right.$

Answer

When $\mathrm{x} \leq 1, \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-1$.
x 0.090 .0990 .9990 .9999
$\mathrm{f}(\mathrm{x})-0.19-0.199-001999-0.00019999$
$\therefore \lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=0$
when $x>1 f(x)=-x^{2}-1$
x 1.11 .011 .00110001
$\mathrm{f}(\mathrm{x})-2.21-2.0201-2.002001-2.00020001$
$\therefore \lim _{x \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})=-2$
Since $\lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x}) \neq \lim _{x \rightarrow 1^{+}} \mathrm{f}(\mathrm{x})$
$\therefore \lim _{x \rightarrow 1} \mathrm{f}(\mathrm{x})$ does not exist.

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