Word Count : 201

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हम दिए गए दो सीमाओं (limits) को हल करते हैं: ### (i) \(\lim_{x\rightarrow+\infty}\left(\frac{x^2}{8x^2-3}\right)^{1/3}\) हल: हम भिन्न (\(\frac{x^2}{8x^2-3}\)) के अंश और हर को \(x^2\) से विभाजित करते हैं: \[ \frac{x^2}{8x^2 - 3} = \frac{1}{8 - \frac{3}{x^2}} \] जब \(x \to +\infty\), तब \(\frac{3}{x^2} \to 0\), इसलिए \[ \frac{1}{8 - 0} = __________ _______ __________ ______ ________.
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