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निम्न दो (02) प्रश्नों के लिए निम्नलिखित पर विचार कीजिए :
माना कि फलन f(x) = x2 + 9 है।
\(\lim_{x \to 0} \frac{\sqrt{f(x)} - 3}{\sqrt{f(x)+7} - 4}\) किसके बराबर है?
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दिया गया है,
फलन \( f(x) = \sqrt{x^2 + 9} - 3 \) और \( g(x) = \sqrt{x^2 + 16} - 4 \).है,
हमें यह ज्ञात करना है:
\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)
अंश और हर दोनों को उनके संगत संयुग्मों से गुणा करने पर:
\( \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4} \)
अंश को सरल करने पर:
\( (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2 \)
हर को सरल करने पर:
\( (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2 \)
अब, व्यंजक बन जाता है:
\( \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \)
सरल करने पर और सीमा का मूल्यांकन करने पर:
\( \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \) यह बन जाता है:
\( \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)
इसलिए, सही उत्तर विकल्प 3 है।
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