Question

The value of $\underset{x\to 0}{lim}\frac{sinx-x}{x^3}$ using taylor series is?

• A. $-\frac{1}{6}$
• B. $\frac{1}{6}$
• C. $-\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{6}$

${lim}_{x\to 0}\frac{sinx-x}{x^3}=?$. We have to find the limit value using Taylor series.
The Taylor series of $sinx$ is $sinx\approx x-\frac{1}{6}{x}^3+...$.
The Taylor series of $x$ is $x\approx x$
The Taylor series of $x^3$ is $x^3\approx x^3$
Hence ${lim}_{x\to 0}\frac{sinx-x}{x^3}\approx {lim}_{x\to 0}\frac{x-\frac{1}{6}{x}^3+...-x}{x^3}$
$={lim}_{x\to 0}\frac{-\frac{1}{6}{x}^3}{x^3}$
$={lim}_{x\to 0}-\frac{1}{6}$
Therefore ${lim}_{x\to 0}\frac{sinx-x}{x^3}=-\frac{1}{6}$