lim_(x->∞)(2x^2+x-3)/(3x^2-4x+1)=? 

প্রশ্ন: \( \lim_{x \to \infty} \frac{2x^2 + x - 3}{3x^2 - 4x + 1} = ? \) 🤔

সমাধান:

আমরা \( x^2 \) দিয়ে লব ও হরকে ভাগ করি। ➗

\( \lim_{x \to \infty} \frac{2x^2 + x - 3}{3x^2 - 4x + 1} = \lim_{x \to \infty} \frac{\frac{2x^2}{x^2} + \frac{x}{x^2} - \frac{3}{x^2}}{\frac{3x^2}{x^2} - \frac{4x}{x^2} + \frac{1}{x^2}} \)

\( = \lim_{x \to \infty} \frac{2 + \frac{1}{x} - \frac{3}{x^2}}{3 - \frac{4}{x} + \frac{1}{x^2}} \)

যেহেতু \( x \to \infty \), তাই \( \frac{1}{x} \to 0 \) এবং \( \frac{1}{x^2} \to 0 \)। 📉

সুতরাং, \( \lim_{x \to \infty} \frac{2 + \frac{1}{x} - \frac{3}{x^2}}{3 - \frac{4}{x} + \frac{1}{x^2}} = \frac{2 + 0 - 0}{3 - 0 + 0} = \frac{2}{3} \)

অতএব, উত্তর: \( \frac{2}{3} \) ✅