Given that f(x) = begin{cases} frac{1-cos 4x} | vedclass

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(B) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal: $

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) For the function $f(x)$ to be continuous at $x = 0$,the left-hand limit,right-hand limit,and the value of the function at $x = 0$ must be equal: $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = a$. First,calculate the left-hand limit: $\lim_{x \rightarrow 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x \rightarrow 0^-} \frac{2 \sin^2(2x)}{x^2} = \lim_{x \rightarrow 0^-} 2 \cdot \left(\frac{\sin 2x}{2x}\right)^2 \cdot 4 = 2 \cdot 1^2 \cdot 4 = 8$. Next,calculate the right-hand limit: $\lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$. Multiply the numerator and denominator by the conjugate $\sqrt{16+\sqrt{x}}+4$: $\lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(16+\sqrt{x})-16} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x \rightarrow 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16+0}+4 = 4+4 = 8$. Since both limits are equal to $8$,for the function to be continuous,$a$ must be $8$.

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

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