Question

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

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

Answer 1

Answer: (B)

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

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx}{x^2}$ $\left(\frac{0}{0}\right)$ form

$=\underset{x\to 0}{lim}\frac{-a\sin ax+b\sin bx}{2x}$ $\left(\frac{0}{0}\right)$ form (using L'Hospital's rule)

$=\underset{x\to 0}{lim}\frac{-a^2\cos ax+b^2\cos bx}{2}$ (using L'Hospital's rule)
$=\frac{b^2-a^2}{2}$