Question

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

Answer 1

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

$={lim}_{x\to 0}\frac{-2sin\left(\frac{ax+bx}{2}\right)sin\left(\frac{ax-bx}{2}\right)}{-2si{n}^2\frac{cx}{2}}(1-cos2\theta =2si{n}^2\theta )$

$={lim}_{x\to 0}\frac{\left(\frac{a+b}{2}\right)x\times \frac{sin\left(\frac{a+b}{2}\right)x}{\left(\frac{a+b}{2}\right)x}\times \left(\frac{a-b}{2}\right)x\times \frac{sin\left(\frac{a-b}{2}\right)x}{\left(\frac{a-b}{2}\right)x}}{\frac{c^2{x}^2}{4}\times \frac{si{n}^2\frac{cx}{2}}{\frac{c^2{x}^2}{4}}}$

$=\frac{(a+b)(a-b)}{c^2}\times \frac{{lim}_{x\to 0}\frac{sin\left(\frac{a+b}{2}\right)x}{\left(\frac{a+b}{2}\right)x}\times {lim}_{x\to 0}\frac{sin\left(\frac{a-b}{2}\right)x}{\left(\frac{a-b}{2}\right)x}}{{\left({lim}_{x\to 0}\frac{sin\frac{cx}{2}}{\frac{cx}{2}}\right)}^2}$

$=\frac{a^2-b^2}{c^2}\times \frac{1\times 1}{1}[\because {lim}_{\theta \to 0}\frac{sin\theta }{\theta }=1]$

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