Question

Evaluate the limit, $\underset{x\to 0}{lim}\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx}$

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

Answer 1

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

$\underset{x\to 0}{lim}\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx}$ ($\frac{0}{0}$ form)

using L-Hospital rule

$=\underset{x\to 0}{lim}\frac{-a\sin ax+b\sin bx\cos cx+c\sin cx\cos bx}{b\cos bx\sin cx+c\sin bx\cos cx}$ ($\frac{0}{0}$ form)

Again using L-Hospital rule

$=\underset{x\to 0}{lim}\frac{-a^2\cos ax+b\left[b\cos bx\cos cx-c\sin bx\sin cx\right]+c\left[c\cos cx\cos bx-b\sin cx\sin bx\right]}{b\left[-b\sin bx\sin cx+c\cos bx\cos cx\right]+c\left[b\cos bx\cos cx-c\sin bx\sin cx\right]}$

$=\frac{-a^2+b\left(b-0\right)+c\left(c-0\right)}{bc+cb}$

$=\frac{-a^2+b^2+c^2}{2bc}$

$=\frac{b^2+c^2-a^2}{2bc}$