Question

For which value of $a$ and $b$, $\underset{x\to 0}{lim}\frac{\sin 2x}{x^3}+a+\frac{b}{x^2}=0$

• A. $b=-2$ and $a=\frac{4}{3}$
• B. $b=2$ and $a=\frac{4}{3}$
• C. $b=-2$ and $a=\frac{-4}{3}$
• D. None of these

Answer 1

The correct option is A $b=-2$ and $a=\frac{4}{3}$

${lim}_{x\to 0}\frac{sin2x}{x^3}+a+\frac{b}{x^2}=0Usingexpansionofsinwehave,sin2x=2x-\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^5}{5!}-...{lim}_{x\to 0}\frac{2x-\frac{{2x}^3}{3!}+\frac{{2x}^5}{5!}-...}{x^3}+a+\frac{b}{x^2}=0{lim}_{x\to 0}\frac{2}{x^2}-\frac{8}{3!}\frac{32x^2}{5!}-..+a+\frac{b}{x^2}=0Forlimittobefinite,\frac{b}{x^2}-\frac{2}{x^2}=0\Rightarrow b=2Comparingconstanttermonbothsides:a=\frac{8}{6}=\frac{4}{3}$