Question:
If \( f: \mathbb{R} \to \mathbb{R} \) is defined by
\[
f(x) =
\begin{cases}
\dfrac{2 \sin x - \sin 2x}{2x \cos x}, & \text{if } x \ne 0 \\
a, & \text{if } x = 0
\end{cases}
\]
then the value of \( a \) so that \( f \) is continuous at \( 0 \) is
If \( f: \mathbb{R} \to \mathbb{R} \) is defined by
\[
f(x) =
\begin{cases}
\dfrac{2 \sin x - \sin 2x}{2x \cos x}, & \text{if } x \ne 0 \\
a, & \text{if } x = 0
\end{cases}
\]
then the value of \( a \) so that \( f \) is continuous at \( 0 \) is
Show Hint
Use L'Hopital's Rule for $0/0$ or $\infty/\infty$ limit forms.
Updated On: Apr 10, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Continuity Condition
For continuity at $x=0$, $f(0) = \lim_{x \to 0} f(x)$. So $a = \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{2x \cos x}$.
Step 2: Evaluate Limit
Applying L'Hopital's Rule ($0/0$ form): $a = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2x}{2(\cos x - x \sin x)}$. $a = \frac{2(1) - 2(1)}{2(1 - 0)} = \frac{0}{2} = 0$.
Final Answer: (d)
For continuity at $x=0$, $f(0) = \lim_{x \to 0} f(x)$. So $a = \lim_{x \to 0} \frac{2 \sin x - \sin 2x}{2x \cos x}$.
Step 2: Evaluate Limit
Applying L'Hopital's Rule ($0/0$ form): $a = \lim_{x \to 0} \frac{2 \cos x - 2 \cos 2x}{2(\cos x - x \sin x)}$. $a = \frac{2(1) - 2(1)}{2(1 - 0)} = \frac{0}{2} = 0$.
Final Answer: (d)
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