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(D) 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,i.

Vedclass · Accepted Answer

$8$

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(D) 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,i.e.,$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0) = a$. $1$. Left-hand limit $(LHL)$: $\lim_{x 	o 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x 	o 0^-} \frac{2\sin^2(2x)}{x^2} = \lim_{x 	o 0^-} 2 \left( \frac{\sin 2x}{2x} ight)^2 	imes 4 = 2 	imes 1^2 	imes 4 = 8$. $2$. Right-hand limit $(RHL)$: $\lim_{x 	o 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$. Rationalizing the denominator: $\lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x 	o 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x 	o 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16+0}+4 = 4+4 = 8$. Since $LHL$ = $RHL$ = $8$,for the function to be continuous,$a$ must be equal to $8$.

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

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