Question:
What is \(\lim_{h \to 0} \frac{\sqrt{2x+3h} – \sqrt{2x}}{2h}\) equal to?
\(\lim_{h \to 0} \frac{\sqrt{2x+3h} – \sqrt{2x}}{2h}\) किसके बराबर है?
Options:
- \(\frac{1}{2\sqrt{2x}}\)
- \(\frac{\sqrt{2x}}{3}\)
- \(\frac{3}{2\sqrt{2x}}\)
- \(\frac{3}{4\sqrt{2x}}\)
- \(\frac{1}{2\sqrt{2x}}\)
- \(\frac{\sqrt{2x}}{3}\)
- \(\frac{3}{2\sqrt{2x}}\)
- \(\frac{3}{4\sqrt{2x}}\)
Correct Answer: d)
Explanation:
Rationalizing the numerator:
\[\frac{\sqrt{2x+3h} – \sqrt{2x}}{2h} = \frac{(2x + 3h) – 2x}{2h(\sqrt{2x+3h} + \sqrt{2x})} = \frac{3h}{2h(\sqrt{2x+3h} + \sqrt{2x})} = \frac{3}{2(\sqrt{2x+3h} + \sqrt{2x})}\]
Taking the limit as \(h \to 0\):
\[\lim_{h \to 0} \frac{3}{2(\sqrt{2x} + \sqrt{2x})} = \frac{3}{2(2\sqrt{2x})} = \frac{3}{4\sqrt{2x}}\]
Hence, the correct option is (d).
अंश का परिमेयकरण करने पर:
\[\frac{3h}{2h(\sqrt{2x+3h} + \sqrt{2x})} = \frac{3}{2(\sqrt{2x+3h} + \sqrt{2x})}\]
\(h \to 0\) सीमा लेने पर:
\[\frac{3}{2(2\sqrt{2x})} = \frac{3}{4\sqrt{2x}}\]
अतः सही विकल्प (d) है।