Question

If $f\left(x\right)=\frac{2\sin x-\sin 2x}{x^3}$ where $x\ne 0$, then $\underset{x\to 0}{lim}f\left(x\right)$ has the value;

• A. $0$
• B. $1$
• C. $2$
• D. Not defined.

Answer 1

Answer: (B)

$1$

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\left(\frac{2sinx-sin2x}{x^3}\right)={lim}_{x\to 0}\left(\frac{2[x-(\frac{x^3}{3!})+(\frac{x^5}{5!})-....]-\left[\right(2x)-(\frac{2x^3}{3!})+(\frac{2x^5}{5!})+....]}{x^3}\right)={lim}_{x\to 0}\left(\frac{x63\left(\right(\frac{-1}{3})+(\frac{4}{3}\left)\right)+x^5(....)+higherterm}{x^3}\right)=\left(\frac{-1}{3}\right)+\left(\frac{4}{3}\right)=\left(\frac{3}{3}\right)=1$