Question:
If
\[
f(x)=
\begin{cases}
\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2}, & x\neq \pi\\
k, & x=\pi
\end{cases}
\]
is continuous at \(x=\pi\), then \(k=\)
If
\[
f(x)=
\begin{cases}
\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2}, & x\neq \pi\\
k, & x=\pi
\end{cases}
\]
is continuous at \(x=\pi\), then \(k=\)
Show Hint
Whenever radicals appear in limits, rationalization is usually the first step.
Also remember:
\[
1+\cos x = 2\cos^2\frac{x}{2}
\]
and
\[
\lim_{t\to0}\frac{\sin t}{t}=1
\]
Updated On: Jun 17, 2026
- \(1\)
- \(\dfrac12\)
- \(2\)
- \(\dfrac14\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For continuity at \(x=a\),
\[
\lim_{x\to a}f(x)=f(a)
\]
Hence, for the function to be continuous at \(x=\pi\),
\[
k=\lim_{x\to \pi}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^2}
\]
This limit is evaluated using trigonometric identities and rationalization.
Step 1: Rationalize the numerator.
Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} \cdot \frac{\sqrt{2+\cos x}+1}{\sqrt{2+\cos x}+1} \] Then, \[ = \frac{2+\cos x-1}{(\pi-x)^2(\sqrt{2+\cos x}+1)} \] \[ = \frac{1+\cos x}{(\pi-x)^2(\sqrt{2+\cos x}+1)} \]
Step 2: Use the identity for \(1+\cos x\).
Recall: \[ 1+\cos x = 2\cos^2\frac{x}{2} \] Hence, \[ = \frac{2\cos^2(x/2)} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \] Now write: \[ \cos\frac{x}{2} = \sin\left(\frac{\pi-x}{2}\right) \] Thus, \[ = \frac{2\sin^2\left(\frac{\pi-x}{2}\right)} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \]
Step 3: Apply standard limit.
As \(x\to\pi\), \[ \frac{\sin\left(\frac{\pi-x}{2}\right)} {\frac{\pi-x}{2}} \to 1 \] Therefore, \[ 2\sin^2\left(\frac{\pi-x}{2}\right) \sim 2\left(\frac{\pi-x}{2}\right)^2 = \frac{(\pi-x)^2}{2} \] Hence, \[ \lim_{x\to\pi} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} = \lim_{x\to\pi} \frac{\frac{(\pi-x)^2}{2}} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \] \[ = \frac1{2(1+1)} = \frac14 \] Thus, \[ k=\boxed{\frac14} \]
Step 1: Rationalize the numerator.
Multiply numerator and denominator by the conjugate: \[ \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} \cdot \frac{\sqrt{2+\cos x}+1}{\sqrt{2+\cos x}+1} \] Then, \[ = \frac{2+\cos x-1}{(\pi-x)^2(\sqrt{2+\cos x}+1)} \] \[ = \frac{1+\cos x}{(\pi-x)^2(\sqrt{2+\cos x}+1)} \]
Step 2: Use the identity for \(1+\cos x\).
Recall: \[ 1+\cos x = 2\cos^2\frac{x}{2} \] Hence, \[ = \frac{2\cos^2(x/2)} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \] Now write: \[ \cos\frac{x}{2} = \sin\left(\frac{\pi-x}{2}\right) \] Thus, \[ = \frac{2\sin^2\left(\frac{\pi-x}{2}\right)} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \]
Step 3: Apply standard limit.
As \(x\to\pi\), \[ \frac{\sin\left(\frac{\pi-x}{2}\right)} {\frac{\pi-x}{2}} \to 1 \] Therefore, \[ 2\sin^2\left(\frac{\pi-x}{2}\right) \sim 2\left(\frac{\pi-x}{2}\right)^2 = \frac{(\pi-x)^2}{2} \] Hence, \[ \lim_{x\to\pi} \frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} = \lim_{x\to\pi} \frac{\frac{(\pi-x)^2}{2}} {(\pi-x)^2(\sqrt{2+\cos x}+1)} \] \[ = \frac1{2(1+1)} = \frac14 \] Thus, \[ k=\boxed{\frac14} \]
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