If L = lim_{x^2 to a} frac{b - cos(x^2 - a)}{ | vedclass

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(B) Let $t = x^2 - a$. As $x^2 	o a$,$t 	o 0$.The limit becomes $L = \lim_{t 	o 0} \frac{b - \cos(t)}{t \sin(ct)}$.For the limit to be finite and n

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$L = \frac{1}{2}, b = 1, c = 1$

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(B) Let $t = x^2 - a$. As $x^2 	o a$,$t 	o 0$.The limit becomes $L = \lim_{t 	o 0} \frac{b - \cos(t)}{t \sin(ct)}$.For the limit to be finite and non-zero,the numerator must approach $0$ as $t 	o 0$ because the denominator approaches $0$. Thus,$b - \cos(0) = 0 \implies b = 1$.Now,$L = \lim_{t 	o 0} \frac{1 - \cos(t)}{t \sin(ct)} = \lim_{t 	o 0} \frac{2 \sin^2(t/2)}{t \sin(ct)}$.Using the standard limit $\lim_{u 	o 0} \frac{\sin u}{u} = 1$,we multiply and divide by $t$ and $ct$:$L = \lim_{t 	o 0} \frac{2 \sin^2(t/2)}{t \cdot (ct) \cdot \frac{\sin(ct)}{ct}} = \lim_{t 	o 0} \frac{2 (t/2)^2}{t \cdot ct} = \lim_{t 	o 0} \frac{2 \cdot \frac{t^2}{4}}{ct^2} = \frac{1/2}{c}$.Since $L$ is non-zero and finite,and comparing with the options,we find $c = 1$ and $L = 1/2$.

If $L = \lim_{x^2 \to a} \frac{b - \cos(x^2 - a)}{(x^2 - a) \sin(c(x^2 - a))}$ is a non-zero finite value $(a > 0)$,then:

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