Question

Lim x tends to 0 {cos(ax)- cos(bx)}/x^2

Answer 1

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Answer 2

So, The answer is $\frac{b^{2} -a^{2} }{2}$

Step-by-step explanation:

Given,

$\lim_{x \to 0} \frac{cos(ax)-cos(bx)}{x^{2} } =?$

∴ $\frac{cos(ax)-cos(bx)}{x^{2} } =\frac{2sin\frac{(ax+bx)}{2}sin\frac{(bx-ax)}{2} }{x^{2} }$

                        $=2\frac{sin.x\frac{a+b}{2} }{x.\frac{a+b}{2} } \frac{a+b}{2}\times \frac{sin.x\frac{b-a}{2} }{x.\frac{b-a}{2} } \frac{b-a}{2}$

∴ $\lim_{x \to 0} \frac{cos(ax)-cos(bx)}{x^{2} } =2 \lim_{x \to 0} \frac{sin.x\frac{a+b}{2} }{x.\frac{a+b}{2} } \frac{a+b}{2}\times \lim_{x \to 0} \frac{sin.x\frac{b-a}{2} }{x.\frac{b-a}{2} } \frac{b-a}{2}$

                                    $=2\times \frac{a+b}{2}\times \frac{b-a}{2}$  (∵ $\lim_{x \to 0} \frac{sinx}{x} =1$ )

                                    $=\frac{b^{2} -a^{2} }{2}$ (∵ $(a+b)(a-b)=a^{2} -b^{2}$ )

∴ The answer is $\frac{b^{2} -a^{2} }{2}$