Question

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}=$

• A. $0$
• B. $1$
• C. $\frac{a^2-b^2}{2}$
• D. $\frac{b^2-a^2}{2}$

Answer 1

Answer: (D)

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

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}$
$=\underset{x\to 0}{lim}\frac{2\sin \frac{(a+b)x}{2}\sin \frac{(b-a)x}{2}}{x^2}$
$=\frac{2(a+b)(b-a)}{4}\underset{x\to 0}{lim}\frac{\sin \frac{(a+b)x}{2}\sin \frac{(b-a)x}{2}}{\frac{(a+b)x}{2}\frac{(b-a)x}{2}}$
$=\frac{b^2-a^2}{2}$
Hence, option 'D' is correct.