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(B) For a function to be continuous at $x=0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.Given $f(0) = k$.We calculate $\lim_{x \righta

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$5$

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(B) For a function to be continuous at $x=0$,the limit of $f(x)$ as $x \rightarrow 0$ must equal $f(0)$.Given $f(0) = k$.We calculate $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} \frac{1+3x^2-\cos 2x}{x^2}$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we substitute:$\lim_{x \rightarrow 0} \frac{1+3x^2-(1-2\sin^2 x)}{x^2} = \lim_{x \rightarrow 0} \frac{1+3x^2-1+2\sin^2 x}{x^2}$.$= \lim_{x \rightarrow 0} \frac{3x^2+2\sin^2 x}{x^2} = \lim_{x \rightarrow 0} \left( 3 + 2\left(\frac{\sin x}{x}\right)^2 \right)$.Since $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:$= 3 + 2(1)^2 = 3 + 2 = 5$.Thus,for continuity,$k = 5$.

If $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{1+3x^2-\cos 2x}{x^2}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then $k$ is equal to

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