If f(x) = begin{cases} frac{cos(ax) - cos(bx) | vedclass

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(B) To check the continuity of $f(x)$ at $x = 0$,we need to evaluate $\lim_{x \rightarrow 0} f(x)$.Given $f(x) = \frac{\cos(ax) - \cos(bx)}{x^2}$ for

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$f$ is continuous at $x = 0$

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(B) To check the continuity of $f(x)$ at $x = 0$,we need to evaluate $\lim_{x ightarrow 0} f(x)$.Given $f(x) = \frac{\cos(ax) - \cos(bx)}{x^2}$ for $x 
eq 0$.Using the limit formula $\lim_{x ightarrow 0} \frac{1 - \cos(	heta)}{x^2} = \frac{	heta^2}{2x^2} = \frac{	heta^2}{2}$,we have:$\lim_{x ightarrow 0} \frac{\cos(ax) - \cos(bx)}{x^2} = \lim_{x ightarrow 0} \left[ \frac{1 - \cos(bx)}{x^2} - \frac{1 - \cos(ax)}{x^2} ight]$$= \lim_{x ightarrow 0} \left[ \frac{2 \sin^2(bx/2)}{x^2} - \frac{2 \sin^2(ax/2)}{x^2} ight]$$= \lim_{x ightarrow 0} \left[ 2 \left( \frac{\sin(bx/2)}{x} ight)^2 - 2 \left( \frac{\sin(ax/2)}{x} ight)^2 ight]$$= 2 \left( \frac{b}{2} ight)^2 - 2 \left( \frac{a}{2} ight)^2 = 2 \left( \frac{b^2}{4} - \frac{a^2}{4} ight) = \frac{b^2 - a^2}{2}$.Since $\lim_{x ightarrow 0} f(x) = \frac{b^2 - a^2}{2}$ and $f(0) = \frac{b^2 - a^2}{2}$,the function $f(x)$ is continuous at $x = 0$.

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

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