MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = $
  • A
    $\frac{{{a^2} - {b^2}}}{2}$
  • $\frac{{{b^2} - {a^2}}}{2}$
  • C
    ${a^2} - {b^2}$
  • D
    ${b^2} - {a^2}$

Answer

Correct option: B.
$\frac{{{b^2} - {a^2}}}{2}$
b
(b) $\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{2\,\sin \,\left( {\frac{{a + b}}{2}} \right)x\,.\,\sin \,\left( {\frac{{b - a}}{2}} \right)\,x}}{{\left( {\frac{{a + b}}{2}} \right)x\,.\frac{2}{{a + b}}.\frac{2}{{b - a}}.\left( {\frac{{b - a}}{2}} \right)x}} = \frac{{{b^2} - {a^2}}}{2}$

Aliter : Apply  $ L$-Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \,a\sin ax + b\sin bx}}{{2x}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \,{a^2}\cos ax + {b^2}\cos bx}}{2} = \frac{{{b^2} - {a^2}}}{2}.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free