A) \[\underset{x\to a}{\mathop{\lim }}\,f(x)=a\]
B) \[f(x)\]is continuous at\[x=a\]
C) \[f(x)\]is discontinuous at\[x=a\]
D) None of these
Correct Answer: B
Solution :
\[f(a)=0\] \[\underset{x\to a-}{\mathop{\lim }}\,\,f(x)=\underset{x\to a-}{\mathop{\lim }}\,\left( \frac{{{x}^{2}}}{a}-a \right)=\underset{h\to 0}{\mathop{\lim }}\,\,\left\{ \frac{{{(a-h)}^{2}}}{a}-a \right\}=0\] and \[\underset{x\to a+}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\left\{ a-\frac{{{(a+h)}^{2}}}{a} \right\}=0\] Hence it is continuous at \[x=a\].
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