Question

$\displaystyle \lim_{x \to 0}$ $\frac{cos\,ax-cos\,bx}{cos\,cx-1}=$

• A. $a^{2}+b^{2}-c^{2}$
• B. $\left(a^{2}+b^{2}\right)/c^{2}$
• C. $a^{2}+b^{2}+c^{2}$
• D. $\left(a^{2}-b^{2}\right)/c^{2}$

Answer 1

The Correct Option is D

We have $\displaystyle \lim_{x \to 0}$ $\frac{-2\,sin\left(\frac{\left(a+b\right)}{2}x\right)sin\left(\frac{\left(a-b\right)}{2}x\right)}{-2\,sin^{2}\left(cx/2\right)}$ $=\displaystyle \lim_{x \to 0}$ $\frac{sin\left(\frac{\left(a+b\right)x}{2}\right)\cdot sin\left(\frac{\left(a-b\right)x}{2}\right)}{x^{2}}\cdot\frac{x^{2}}{sin^{2} \, \frac{cx}{2}}$ $=\displaystyle \lim_{x \to 0}$ $\frac{sin\frac{\left(a+b\right)x}{2}}{\frac{\left(a+b\right)x}{2}\cdot\left(\frac{2}{a+b}\right)}\cdot\frac{sin \frac{\left(a-b\right)x}{2}}{\frac{\left(a-b\right)x}{2}\cdot\frac{2}{a-b}}\cdot\frac{\left(\frac{cx}{2}\right)^{2} \times \frac{4}{c^{2}}}{sin^{2}\, \frac{cx}{2}}$ $=\bigg[\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right)\left(\frac{4}{c^{2}}\right) \displaystyle \lim_{x \to 0}\left\{\frac{sin \frac{\left(a+b\right)x}{2}}{\frac{\left(a+b\right)x}{2}}\right\} \times$ $\displaystyle \lim_{x \to 0}$$\left\{\frac{sin \frac{\left(a-b\right)x}{2}}{\left(\frac{a-b}{2}\right)x}\right\} $$\displaystyle \lim_{x \to 0}$$\left\{\frac{\frac{cx}{2}}{sin \frac{cx}{2}}\right\}^{2}\bigg]$ $=\left(\frac{a+b}{2} \times \frac{a-b}{2} \times \frac{4}{c^{2}}\right)$ $=\frac{a^{2}-b^{2}}{c^{2}}$