Question

Solve $\underset{x\to 0}{Lt}\frac{\cos ax-\cos bx}{x^2}$

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

Answer 1

The correct option is B $\frac{b^2-a^2}{2}$

${lim}_{x\to 0}\frac{cosax-cosbx}{x^2}=\frac{0}{0}$ Form.

${lim}_{x\to 0}\frac{-asinax+bsinbx}{2x}=\frac{0}{0}$ Form.

${lim}_{x\to 0}\frac{-a^{2}cosax+b^{2}cosbx}{2}=\frac{b^2-a^2}{2}$