Given 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 condition $\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$ must hold.First,calculate the left-hand

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$8$

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(A) For $f(x)$ to be continuous at $x=0$,the condition $\lim_{x 	o 0^-} f(x) = f(0) = \lim_{x 	o 0^+} f(x)$ must hold.First,calculate the left-hand limit $(LHL)$:$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x 	o 0^-} \frac{2\sin^2(2x)}{x^2} = 2 \lim_{x 	o 0^-} \left(\frac{\sin 2x}{2x}ight)^2 	imes 4 = 2 	imes 1^2 	imes 4 = 8$.Next,calculate the right-hand limit $(RHL)$:$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$.Rationalizing the denominator:$\lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x 	o 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$.Since $	ext{LHL} = 	ext{RHL} = 8$,for continuity,$a = 8$.

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

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