Solveeit Logo
Login

Question

Mathematics Question on Limits

lim⁡x→∞(x23x−2−x3)=\displaystyle\lim_{x \to\infty} \left(\frac{x^{2}}{3x-2}-\frac{x}{3}\right)=

A

13\frac{1}{3}

B

23\frac{2}{3}

C

−23\frac{-2}{3}

D

29\frac{2}{9}

Answer

29\frac{2}{9}

Explanation

Solution

Consider lim⁡x→∞[x23x−2−x3]\displaystyle \lim_{x \to\infty }\left[\frac{x^{2}}{3x-2}-\frac{x}{3}\right]  =limx→∞[3x2−x(3x−2)3(3x−2)]\displaystyle\ =lim_{x \to\infty } \left [\frac{3x^{2}-x\left(3x-2\right)}{3\left(3x-2\right)}\right] =lim⁡x→∞2x3(3x−2)=lim⁡x→∞2x3x[3−2x]=\displaystyle \lim _{x\to\infty} \frac{2x}{3\left(3x-2\right)} =\displaystyle\lim_{x\to\infty} \frac{2x}{3x\left[3-\frac{2}{x}\right]} =lim⁡x→∞2313−2x=23×13−0=29=\displaystyle \lim _{x\to\infty} \frac{2}{3} \frac{1}{3-\frac{2}{x}}=\frac{2}{3} \times\frac{1}{3-0}=\frac{2}{9}