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If $f(x)=\left\{\begin{array}{rr}\left(x^2 / a\right)-a, & \text { when } x \lt a \\ 0, & \text { when } x=a, \\ a-\left(x^2 / a\right), & \text {when } x \gt a\end{array}\right.$ then
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The correct answer is:
$f(x)$ is continuous at $x=a$
$f(a)=0$
$\lim _{x \rightarrow a-} f(x)=\lim _{x \rightarrow a}^{-}\left(\frac{x^2}{a}-a\right)=\lim _{h \rightarrow 0}\left\{\frac{(a-h)^2}{a}-a\right\}=0$
and $\lim _{x \rightarrow a}^{+} f(x)=\lim _{h \rightarrow 0}\left\{a-\frac{(a+h)^2}{a}\right\}=0$
Hence it is continuous at $x=a$
$\lim _{x \rightarrow a-} f(x)=\lim _{x \rightarrow a}^{-}\left(\frac{x^2}{a}-a\right)=\lim _{h \rightarrow 0}\left\{\frac{(a-h)^2}{a}-a\right\}=0$
and $\lim _{x \rightarrow a}^{+} f(x)=\lim _{h \rightarrow 0}\left\{a-\frac{(a+h)^2}{a}\right\}=0$
Hence it is continuous at $x=a$
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