Question

Solve $\underset{x\to 0}{lim}\frac{{\sin }^{2}ax}{{\tan }^{2}bx}=$

• A. $\frac{a}{b}$
• B. $\frac{a^2}{b^2}$
• C. $\frac{b^2}{a^2}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B $\frac{a^2}{b^2}$

$\underset{x\to 0}{lim}\frac{{\left(\frac{\sin ax}{ax}\times ax\right)}^2}{{\left(\frac{\tan bx}{bx}\times bx\right)}^2}$

$\underset{x\to 0}{lim}\frac{a^2{x}^2{\left(\frac{\sin ax}{ax}\right)}^2}{b^2{x}^2{\left(\frac{\tan bx}{bx}\right)}^2}$

$\frac{a^2}{b^2}\left[\underset{x\to 0}{lim}\frac{\sin x}{x}=1,\underset{x\to 0}{lim}\frac{\tan x}{x}=1\right]$

$\underset{x\to 0}{lim}\frac{a^2{x}^2}{b^2{x}^2}=\underset{x\to 0}{lim}\frac{a^2}{b^2}$

Substitute the limit

$\frac{a^2}{b^2}$

Hence, the correct option is $b$