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For \[f(x)=\begin{cases}x^3+3,&x\ne0\\1,&x=0\end{cases}\], which statement is correct at \[x=0\]?

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Question

For \[f(x)=\begin{cases}x^3+3,&x\ne0\\1,&x=0\end{cases}\], which statement is correct at \[x=0\]?

Options

  • \[f(0)=3\] and \[\lim_{x\to0}f(x)=3\], so \[f\] is continuous.

  • \[f(0)=1\] and \[\lim_{x\to0}f(x)=3\], so \[f\] is discontinuous.

  • \[f(0)\] is not defined and \[\lim_{x\to0}f(x)=3\].

  • \[f(0)=1\] and \[\lim_{x\to0}f(x)=1\], so \[f\] is continuous.

MCQ
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Solution

The function value is specified as \[f(0)=1\]. For \[x\ne0\], the expression approaches \[0^3+3=3\], which is not equal to \[f(0)\].

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