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

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Question

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

Vedclass · Accepted Answer

$-4, 3$

Vedclass · Answer

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

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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