If lim _{x rightarrow 0}left(frac{cos 4 x+a c | vedclass

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(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x \rightarrow 0$,the numerator must approach $0$ as $x \rightarrow 0$.Using t

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$-4, 3$

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(C) Let $f(x) = \cos 4x + a \cos 2x + b$. For the limit to be finite as $x ightarrow 0$,the numerator must approach $0$ as $x ightarrow 0$.Using the Taylor series expansion: $\cos 	heta = 1 - \frac{	heta^2}{2!} + \frac{	heta^4}{4!} - \dots$$\cos 4x = 1 - \frac{(4x)^2}{2} + \frac{(4x)^4}{24} - \dots = 1 - 8x^2 + \frac{32}{3}x^4 - \dots$$a \cos 2x = a(1 - \frac{(2x)^2}{2} + \frac{(2x)^4}{24} - \dots) = a - 2ax^2 + \frac{2a}{3}x^4 - \dots$Substituting these into the expression:$f(x) = (1 + a + b) - (8 + 2a)x^2 + (\frac{32}{3} + \frac{2a}{3})x^4 + \dots$For the limit to be finite,the coefficients of $x^0$ and $x^2$ must be zero.$1 + a + b = 0$ and $8 + 2a = 0$.From $8 + 2a = 0$,we get $a = -4$.Substituting $a = -4$ into $1 + a + b = 0$,we get $1 - 4 + b = 0$,which gives $b = 3$.Thus,the values are $a = -4$ and $b = 3$.

If $\lim _{x \rightarrow 0}\left(\frac{\cos 4 x+a \cos 2 x+b}{x^4}\right)$ is finite,then the values of $a, b$ are respectively :

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