Question

$\underset{x\to 0}{\lim }$ $\frac{cosax-cosbx}{coscx-1}=$

• 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

Answer: (D)

We have $\underset{x\to 0}{\lim }$ $\frac{-2sin\left(\frac{\left(a+b\right)}{2}x\right)sin\left(\frac{\left(a-b\right)}{2}x\right)}{-2si{n}^2\left(cx/2\right)}$
$=\underset{x\to 0}{\lim }$ $\frac{sin\left(\frac{\left(a+b\right)x}{2}\right)\cdot sin\left(\frac{\left(a-b\right)x}{2}\right)}{x^2}\cdot \frac{x^2}{si{n}^2\frac{cx}{2}}$
$=\underset{x\to 0}{\lim }$ $\frac{sin\frac{\left(a+b\right)x}{2}}{\frac{\left(a+b\right)x}{2}\cdot \left(\frac{2}{a+b}\right)}\cdot \frac{sin\frac{\left(a-b\right)x}{2}}{\frac{\left(a-b\right)x}{2}\cdot \frac{2}{a-b}}\cdot \frac{{\left(\frac{cx}{2}\right)}^2\times \frac{4}{c^2}}{si{n}^2\frac{cx}{2}}$
$=[\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right)\left(\frac{4}{c^2}\right)\underset{x\to 0}{\lim }\left\{\frac{sin\frac{\left(a+b\right)x}{2}}{\frac{\left(a+b\right)x}{2}}\right\}\times$
$\underset{x\to 0}{\lim }$$\left\{\frac{sin\frac{\left(a-b\right)x}{2}}{\left(\frac{a-b}{2}\right)x}\right\}$$\underset{x\to 0}{\lim }$${\left\{\frac{\frac{cx}{2}}{sin\frac{cx}{2}}\right\}}^2]$
$=\left(\frac{a+b}{2}\times \frac{a-b}{2}\times \frac{4}{c^2}\right)$
$=\frac{a^2-b^2}{c^2}$