Question

Evaluate: $-2\underset{x\to 0}{lim}\frac{\sin \left(\frac{a+b}{2}\right)x}{x}\underset{x\to 0}{lim}\frac{\sin \left(\frac{a-b}{2}\right)x}{x}$

• A. $\frac{b^2-a^2}{2}$
• B. $\frac{a^2-b^2}{2}$
• C. $\frac{b^2}{2}$
• D. None of these

Answer 1

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

$-2{lim}_{x\to 0}\frac{\sin \left(\frac{a+b}{2}\right)x}{x}{lim}_{x\to 0}\frac{\sin \left(\frac{a-b}{2}\right)x}{x}$

$=-2{lim}_{x\to 0}\frac{\sin \left(\frac{a+b}{2}\right)x}{\left(\frac{a+b}{2}\right)x}\times \left(\frac{a+b}{2}\right)\times {lim}_{x\to 0}\frac{\sin \left(\frac{a-b}{2}\right)x}{\left(\frac{a-b}{2}\right)x}\times \left(\frac{a-b}{2}\right)$

$=-2\times 1\times \frac{a+b}{2}\times \frac{a-b}{2}$

$=-2\times \frac{a^2-b^2}{4}$

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