Define f: R to R by f(x) = begin{cases} frac{ | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = a$. First,calculate the left-hand limi

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$8$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = a$. First,calculate the left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x 	o 0^-} \frac{2 \sin^2 2x}{x^2} = \lim_{x 	o 0^-} 8 \left( \frac{\sin 2x}{2x} ight)^2 = 8(1)^2 = 8$. Next,calculate the right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$. Multiply the numerator and denominator by the conjugate $\sqrt{16+\sqrt{x}}+4$: $\lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(16+\sqrt{x})-16} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x 	o 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$. Since $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = 8$,for the function to be continuous,$a$ must be $8$.

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

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