Question:
What is \(\lim_{\theta \to 0^+} \frac{\sqrt{1-\cos\theta}}{\theta}\) equal to?
\(\lim_{\theta \to 0^+} \frac{\sqrt{1-\cos\theta}}{\theta}\) किसके बराबर है?
Options:
- \(\sqrt{2}\)
- \(2\sqrt{2}\)
- \(\frac{1}{\sqrt{2}}\)
- \(-\frac{1}{2\sqrt{2}}\)
- \(\sqrt{2}\)
- \(2\sqrt{2}\)
- \(\frac{1}{\sqrt{2}}\)
- \(-\frac{1}{2\sqrt{2}}\)
Correct Answer: c)
Explanation:
Using the trigonometric identity \(1 – \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)\):
\[\sqrt{1-\cos\theta} = \sqrt{2}\left|\sin\left(\frac{\theta}{2}\right)\right|\]
Since \(\theta \to 0^+\), \(\frac{\theta}{2} > 0\), so \(\left|\sin(\theta/2)\right| = \sin(\theta/2)\).
\[\lim_{\theta \to 0^+} \frac{\sqrt{2}\sin(\theta/2)}{\theta} = \frac{\sqrt{2}}{2} \lim_{\theta \to 0^+} \frac{\sin(\theta/2)}{\theta/2} = \frac{\sqrt{2}}{2} \times 1 = \frac{1}{\sqrt{2}}\]
Hence, the correct option is (c).
त्रिकोणमितीय सर्वसमिका \(1 – \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)\) का उपयोग करने पर:
\[\sqrt{1-\cos\theta} = \sqrt{2}\left|\sin\left(\frac{\theta}{2}\right)\right|\]
चूँकि \(\theta \to 0^+\), अतः \(\sin(\theta/2) > 0\) होगा:
\[\lim_{\theta \to 0^+} \frac{\sqrt{2}\sin(\theta/2)}{2(\theta/2)} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\]
अतः सही विकल्प (c) है।