MHT CET · Maths · Continuity and Differentiability
If \(f(x)=\left\{\begin{array}{cl}x \sin \frac{1}{x} & , x \neq 0 \ k & , x=0\end{array}\right.\) is continuous at
\(x=0\), then the value of \(k\) is
- A 1
- B \(-1\)
- C 0
- D 2
Answer & Solution
Correct Answer
(C) 0
Step-by-step Solution
Detailed explanation
If function \(f(x)\) is continuous at \(x=0\), then
\(
f(0)=\lim _{x \rightarrow 0} f(x)
\)
Given, \(\quad f(0)=k\)
\(
\begin{array}{ll}
\therefore f(0)=k=\lim _{x \rightarrow 0} x \sin \frac{1}{x} \\
\Rightarrow k=0 \quad\left(\because-1 \leq \sin \frac{1}{x} \leq 1\right)
\end{array}
\)
\(
f(0)=\lim _{x \rightarrow 0} f(x)
\)
Given, \(\quad f(0)=k\)
\(
\begin{array}{ll}
\therefore f(0)=k=\lim _{x \rightarrow 0} x \sin \frac{1}{x} \\
\Rightarrow k=0 \quad\left(\because-1 \leq \sin \frac{1}{x} \leq 1\right)
\end{array}
\)
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