Question

${lim}_{x\to 0}\frac{{\tan }^{3}x-{\sin }^{3}x}{x^5}=$

• A. $\frac{5}{2}$
• B. $\frac{3}{2}$
• C. $\frac{3}{5}$
• D. $\frac{2}{5}$

Answer 1

The correct option is B $\frac{3}{2}$

$\underset{x\to 1}{lim}\frac{ta{n}^{3}x-{\sin }^{3}x}{x^5}$

$\underset{x\to 0}{lim}\frac{{(x+\frac{x^3}{3})}^3-{(x-\frac{x^3}{6})}^3}{x^5}$

$\underset{x\to 0}{lim}\frac{x^3+\frac{x^9}{27}+3x^2\frac{x^3}{3}+3x\frac{x^6}{9}-(x^3-\frac{x^4}{6^3}-3x^2\frac{x^3}{6}+3x^2\frac{x^6}{6^2})}{x^5}$

$1+\frac{1}{2}$

$=\frac{3}{2}$.