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(C) For the function $f(x)$ to be continuous at $x = 0$, we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = b$.First

Vedclass · Accepted Answer

$6$

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(C) For the function $f(x)$ to be continuous at $x = 0$, we must have $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x) = f(0) = b$.First, evaluate the left-hand limit:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{a(1-\cos 2x)}{x^2} = \lim_{x \rightarrow 0^-} \frac{a(2 \sin^2 x)}{x^2} = 2a \lim_{x \rightarrow 0^-} \left(\frac{\sin x}{x}\right)^2 = 2a(1)^2 = 2a$.Thus, $2a = b$.Next, evaluate the right-hand limit:$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}$.Rationalizing the denominator:$\lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{(\sqrt{4+\sqrt{x}}-2)(\sqrt{4+\sqrt{x}}+2)} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{(4+\sqrt{x})-4} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}(\sqrt{4+\sqrt{x}}+2)}{\sqrt{x}} = \lim_{x \rightarrow 0^+} (\sqrt{4+\sqrt{x}}+2) = \sqrt{4}+2 = 4$.So, $b = 4$.Substituting $b = 4$ into $2a = b$, we get $2a = 4$, which implies $a = 2$.Therefore, $a+b = 2+4 = 6$.

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

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