limx→∞ xlogx31+x+x2 equals

limx→∞ logx3+x.3log x2×1x1+2x Using L 'Hospital's rule = limx→∞ 3logx2×1x+6logx×1x2 = limx→∞ 3logx2+6logx2x = limx→∞ 6logx×1x+6x2 Using L' Hospital's rle = limx→∞ 6logx+62x = limx→∞ 61x+02 Using L' Hospital's rule = 6∞2=0

limx→∞ xlogx31+x+x2 equals

limx→∞ logx3+x.3log x2×1x1+2x Using L 'Hospital's rule = limx→∞ 3logx2×1x+6logx×1x2 = limx→∞ 3logx2+6logx2x = limx→∞ 6logx×1x+6x2 Using L' Hospital's rle = limx→∞ 6logx+62x = limx→∞ 61x+02 Using L' Hospital's rule = 6∞2=0

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$\lim_{x \to \infty} \frac{x {(logx)}^{3}}{1 + x + x^{2}}$ equals

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Detailed Solution

$\begin{array}{l} \lim_{x \to \infty} \frac{{(logx)}^{3} + x . 3 {(\log x)}^{2} \times \frac{1}{x}}{1 + 2 x} (Using L' Hospital' s rule) \\ = \lim_{x \to \infty} \frac{3 {(logx)}^{2} \times \frac{1}{x} + 6 (logx) \times \frac{1}{x}}{2} \\ = \lim_{x \to \infty} \frac{3 {(logx)}^{2} + 6 logx}{2 x} \\ = \lim_{x \to \infty} \frac{6 logx \times \frac{1}{x} + \frac{6}{x}}{2} (Using L' Hospital' s rle) \\ = \lim_{x \to \infty} \frac{6 logx + 6}{2 x} \\ = \lim_{x \to \infty} \frac{6 (\frac{1}{x}) + 0}{2} (Using L' Hospital' s rule) \\ = \frac{\frac{6}{\infty}}{2} = 0 \end{array}$