Let f(x) = begin{cases} frac{1 - cos 4x}{x^2} | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the function value at $x = 0$.$\lim_{x \rightarrow 0^-} f(x) = f(0)$Given $f

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$8$

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(A) For $f(x)$ to be continuous at $x = 0$,the left-hand limit must equal the function value at $x = 0$.$\lim_{x ightarrow 0^-} f(x) = f(0)$Given $f(0) = a$.$\lim_{x ightarrow 0^-} \frac{1 - \cos 4x}{x^2} = a$Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$,we have $1 - \cos 4x = 2 \sin^2 2x$.$\lim_{x ightarrow 0^-} \frac{2 \sin^2 2x}{x^2} = a$Multiply and divide by $4$ to use the standard limit $\lim_{	heta ightarrow 0} \frac{\sin 	heta}{	heta} = 1$.$\lim_{x ightarrow 0^-} 2 	imes \left( \frac{\sin 2x}{2x} ight)^2 	imes 4 = a$$2 	imes (1)^2 	imes 4 = a$$a = 8$.

Let $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x} + \sqrt{2}}{\sqrt{16 + \sqrt{x}}}, & x > 0 \end{cases}$
If $f(x)$ is continuous at $x = 0$,then the value of $a$ is

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Let $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x} + \sqrt{2}}{\sqrt{16 + \sqrt{x}}}, & x > 0 \end{cases}$If $f(x)$ is continuous at $x = 0$,then the value of $a$ is