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(B) Given that $f(x)$ is continuous at $x = 0$,we must have $\lim_{x \rightarrow 0} f(x) = f(0) = \lambda$. Calculating the limit: $\lim_{x \rightarro

Vedclass · Accepted Answer

$-4$

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(B) Given that $f(x)$ is continuous at $x = 0$,we must have $\lim_{x ightarrow 0} f(x) = f(0) = \lambda$. Calculating the limit: $\lim_{x ightarrow 0} \frac{\cos 3x - \cos x}{x^2}$ Using the formula $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$: $= \lim_{x ightarrow 0} \frac{-2 \sin(2x) \sin(x)}{x^2}$ $= -2 	imes \lim_{x ightarrow 0} \left( \frac{\sin 2x}{x} ight) 	imes \left( \frac{\sin x}{x} ight)$ $= -2 	imes 2 	imes 1 = -4$. Thus,$\lambda = -4$.

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

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