Question

Evaluate $\underset{x\to 0}{lim}\frac{x-ta{n}^{-1}x}{x^3}$ using series expansion

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{7}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $\frac{1}{3}$

${lim}_{x\to 0}\frac{x-ta{n}^{-1}\left(x\right)}{x^3}=?$
Let us use Maclaurian series expansion to evaluate the limit.
The Maclaurian series of $x$ is $x$.
The Maclaurian series of $ta{n}^{-1}\left(x\right)$ is $x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5$.
The Maclaurian series of $x^3$ is $x^3$.
${lim}_{x\to 0}\frac{x-ta{n}^{-1}\left(x\right)}{x^3}={lim}_{x\to 0}\frac{x-(x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5)}{x^3}$
$={lim}_{x\to 0}\frac{x-x+\frac{1}{3}{x}^3-\frac{1}{5}{x}^5}{x^3}$
$={lim}_{x\to 0}\frac{\frac{1}{3}{x}^3-\frac{1}{5}{x}^5}{x^3}$
$={lim}_{x\to 0}\frac{x^3(\frac{1}{3}-\frac{1}{5}{x}^2)}{x^3}$
$={lim}_{x\to 0}(\frac{1}{3}-\frac{1}{5}{x}^2)$
${lim}_{x\to 0}\frac{x-ta{n}^{-1}\left(x\right)}{x^3}=\frac{1}{3}$