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Q. $\displaystyle \lim_{x \to \infty}$ $\left(\frac{x^{3}}{3x^{2}-4}-\frac{x^{2}}{3x+2}\right)$ is equal to

Limits and Derivatives

Solution:

$\displaystyle \lim_{x \to \infty}$$\left(\frac{x^{3}}{3x^{2}-4}-\frac{x^{2}}{3x+2}\right)$

$=\displaystyle \lim_{x \to \infty}$$\left(\frac{x^{3}\left(3x+2\right)-x^{2}\left(3x^{2}-4\right)}{\left(3x^{2}-4\right)\left(3x+2\right)}\right)$

$=\displaystyle \lim_{x \to \infty}$$\left(\frac{3x^{4}+2x^{3}-3x^{4}+4x^{2}}{9x^{3}+6x^{2}-12x-8}\right)$

$=\displaystyle \lim_{x \to \infty}$$\left(\frac{2x^{3}+4x^{2}}{9x^{3}+6x^{2}-12x-8}\right)$

$=\displaystyle \lim_{x \to \infty}$$\left(\frac{2+\frac{4}{x}}{9+\frac{6}{x}-\frac{12}{x^{2}}-\frac{8}{x^{3}}}\right)$

$=\frac{2}{9}\,\left[\because \displaystyle \lim_{x \to \infty} \frac{a}{x}=0\right]$