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If $f(x)=\left\{\begin{array}{l}x, \text { if } x \text { is rational } \\ -x, \text { if } x \text { is irrational then } \lim _{x \rightarrow 0} f(x) \text { is }\end{array}\right.$
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Equal to 0
$\begin{aligned} & \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0-h)=0 \\ & \text { and } \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0}-(0+h)=0 \\ & \therefore \lim _{x \rightarrow 0} f(x)=0,\left(\because \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)\right) .\end{aligned}$
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