Question

$\underset{x\to 0}{lim}\frac{1-\cos 2x^3}{x^2.{\tan }^{2}3x{\sin }^{2}4x}=$

• A. $\frac{1}{72}$
• B. $\frac{1}{48}$
• C. $\frac{1}{24}$
• D. $\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{72}$

$\underset{x\to 0}{Lim}\frac{\frac{1-cos2x^3}{(2x^3{)}^2}}{\frac{x^{2}ta{n}^{2}3xsi{n}^{2}4x}{(2x^3{)}^2}}$
$\underset{x\to 0}{Lim}\frac{\frac{1}{2}}{\frac{x^2}{x^2}\frac{9ta{n}^2\left(3x\right)}{(3x{)}^2}\frac{4si{n}^{2}4x}{(4x{)}^2}}$
$=\frac{1}{72}$