${\displaystyle f\left(x\right)=\frac{x+2}{x(x+3)}+\frac{3x+1}{x(x+4)}}$ হলে ${\displaystyle \underset{x\to \infty }{lim}xf\left(x\right)=?}$

Explanation:

Another Explanation (5): ```html bài toán: \(\displaystyle f(x)=\frac{x+2}{x(x+3)}+\frac{3x+1}{x(x+4)}\) হলে \(\displaystyle \lim_{x\to \infty} x f(x) = ?\) 🤔 সমাধান: প্রথমে \( f(x) \) কে সরল করি: \(\displaystyle f(x) = \frac{x+2}{x(x+3)} + \frac{3x+1}{x(x+4)}\) \(\displaystyle = \frac{(x+2)(x+4) + (3x+1)(x+3)}{x(x+3)(x+4)}\) \(\displaystyle = \frac{x^2 + 6x + 8 + 3x^2 + 10x + 3}{x(x^2 + 7x + 12)}\) \(\displaystyle = \frac{4x^2 + 16x + 11}{x^3 + 7x^2 + 12x}\) এখন, \(\displaystyle \lim_{x\to \infty} x f(x) \) এর মান বের করি: \(\displaystyle \lim_{x\to \infty} x f(x) = \lim_{x\to \infty} x \cdot \frac{4x^2 + 16x + 11}{x^3 + 7x^2 + 12x}\) \(\displaystyle = \lim_{x\to \infty} \frac{4x^3 + 16x^2 + 11x}{x^3 + 7x^2 + 12x}\) লব ও হরকে \( x^3 \) দিয়ে ভাগ করে পাই: \(\displaystyle = \lim_{x\to \infty} \frac{4 + \frac{16}{x} + \frac{11}{x^2}}{1 + \frac{7}{x} + \frac{12}{x^2}}\) যখন \( x \to \infty \), তখন \(\displaystyle \frac{16}{x} \to 0 \), \(\displaystyle \frac{11}{x^2} \to 0 \), \(\displaystyle \frac{7}{x} \to 0 \) এবং \(\displaystyle \frac{12}{x^2} \to 0 \) হয়। সুতরাং, \(\displaystyle \lim_{x\to \infty} x f(x) = \frac{4 + 0 + 0}{1 + 0 + 0} = \frac{4}{1} = 4\) 🎉 অতএব, \(\displaystyle \lim_{x\to \infty} x f(x) = 4 \) । ✅ ```