lim_(xto0)(root()(a+x^2)-root()(a-x^2))/x^2=? 

প্রশ্ন: \( \lim_{x \to 0} \frac{\sqrt{a+x^2} - \sqrt{a-x^2}}{x^2} = ? \)

উত্তর: \( \frac{1}{\sqrt{a}} \)

সমাধান:

আমরা লিমিটটি নির্ণয় করার জন্য প্রথমে লব ও হরকে \( \sqrt{a+x^2} + \sqrt{a-x^2} \) দিয়ে গুণ করি। 😇

\( \lim_{x \to 0} \frac{\sqrt{a+x^2} - \sqrt{a-x^2}}{x^2} = \lim_{x \to 0} \frac{(\sqrt{a+x^2} - \sqrt{a-x^2})(\sqrt{a+x^2} + \sqrt{a-x^2})}{x^2(\sqrt{a+x^2} + \sqrt{a-x^2})} \)

\( = \lim_{x \to 0} \frac{(a+x^2) - (a-x^2)}{x^2(\sqrt{a+x^2} + \sqrt{a-x^2})} \)

\( = \lim_{x \to 0} \frac{2x^2}{x^2(\sqrt{a+x^2} + \sqrt{a-x^2})} \)

\( = \lim_{x \to 0} \frac{2}{\sqrt{a+x^2} + \sqrt{a-x^2}} \)

এখন, \( x \to 0 \) হলে, 🤔

\( = \frac{2}{\sqrt{a+0^2} + \sqrt{a-0^2}} \)

\( = \frac{2}{\sqrt{a} + \sqrt{a}} \)

\( = \frac{2}{2\sqrt{a}} \)

\( = \frac{1}{\sqrt{a}} \) 🎉

সুতরাং, \( \lim_{x \to 0} \frac{\sqrt{a+x^2} - \sqrt{a-x^2}}{x^2} = \frac{1}{\sqrt{a}} \) ।