Question

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

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

Answer 1

Answer: (B)

$\frac{3}{2}$

$\underset{x\to 0}{lim}\frac{ta{n}^{3}x-si{n}^{3}x}{x^5}$

$=\underset{x\to 0}{lim}\frac{si{n}^{3}x[\frac{1}{co{s}^{3}x}-1]}{x^3\cdot x^2}$

$[since,\underset{x\to 0}{lim}\frac{sinx}{x}=1]$

So, $\Rightarrow \underset{x\to 0}{lim}\frac{si{n}^{3}x[se{c}^{3}x-1]}{x^3\cdot x^2}$

$=\underset{x\to 0}{lim}\frac{se{c}^{3}x-1}{x^2}=\frac{0}{0}form$

Using L Hospital's rule, we get

$=\underset{x\to 0}{lim}\frac{3se{c}^{2}x.secxtanx-0}{2x}$

$=\underset{x\to 0}{lim}\frac{3}{2}\times \frac{1}{co{s}^{4}x}\frac{sinx}{x}$

$=\underset{x\to 0}{lim}\frac{3}{2}\frac{1}{co{s}^{4}x}$

$=\frac{3}{2}$