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Mathematics Question on limits and derivatives

lim⁡x→0\displaystyle \lim_{x \to 0} cos ax−cos bxcos cx−1=\frac{cos\,ax-cos\,bx}{cos\,cx-1}=

A

a2+b2−c2a^{2}+b^{2}-c^{2}

B

(a2+b2)/c2\left(a^{2}+b^{2}\right)/c^{2}

C

a2+b2+c2a^{2}+b^{2}+c^{2}

D

(a2−b2)/c2\left(a^{2}-b^{2}\right)/c^{2}

Answer

(a2−b2)/c2\left(a^{2}-b^{2}\right)/c^{2}

Explanation

Solution

We have lim⁡x→0\displaystyle \lim_{x \to 0} −2 sin((a+b)2x)sin((a−b)2x)−2 sin2(cx/2)\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)} =lim⁡x→0=\displaystyle \lim_{x \to 0} sin((a+b)x2)⋅sin((a−b)x2)x2⋅x2sin2 cx2\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}} =lim⁡x→0=\displaystyle \lim_{x \to 0} sin(a+b)x2(a+b)x2⋅(2a+b)⋅sin(a−b)x2(a−b)x2⋅2a−b⋅(cx2)2×4c2sin2 cx2\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 lim⁡x→0\displaystyle \lim_{x \to 0} \left\\{\frac{sin \frac{\left(a-b\right)x}{2}}{\left(\frac{a-b}{2}\right)x}\right\\} lim⁡x→0\displaystyle \lim_{x \to 0} \left\\{\frac{\frac{cx}{2}}{sin \frac{cx}{2}}\right\\}^{2}\bigg] =(a+b2×a−b2×4c2)=\left(\frac{a+b}{2} \times \frac{a-b}{2} \times \frac{4}{c^{2}}\right) =a2−b2c2=\frac{a^{2}-b^{2}}{c^{2}}