Question

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

Answer 1

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Answer 2

Answer:

The value of $\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}=$$\frac{b^{2}-a^{2}}{2}$

Step-by-step explanation:

A limit is a function which has a variable and its value. Thus as we substitute the limits the variable approaches that value.

 

Given expression: $\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{x^{2}}=\lim _{x \rightarrow 0} \frac{-2 \sin \left(\frac{a x+b x}{3}\right) \sin \left(\frac{a x-b x}{2}\right)}{x^{2}}$

[Since, we know, $\cos p-\cos q=-2 \sin \frac{(p+q)}{2} \sin \frac{(p-q)}{2} =\lim _{x \rightarrow 0} \frac{-2 \sin \left(\frac{a+b}{2}\right) x}{x} \quad \lim _{x \rightarrow 0} \frac{\sin \left(\frac{a-b}{3}\right) x}{x}$

$=-2\left(\frac{a+b}{2}\right)\left(\frac{a-b}{2}\right)$ [Since, we know $\lim _{x\rightarrow 0} \frac{\sin p x}{x}=p]$

$=-2\left(\frac{a^{2}-b^{2}}{4}\right)(p+q)(p-q)=p^{2}-q^{2} ]$

$=-\left(\frac{a^{2}-b^{2}}{2}\right) \quad=\bold{\frac{b^{2}-a^{2}}{2}}$