If begin{aligned} f(x) &= frac{4 sin pi x}{5 | vedclass

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(A) Since $f(x)$ is continuous at $x = 0$,the limit of the function as $x 	o 0$ must equal the value of the function at $x = 0$. $\lim_{x 	o 0} f(x)

Vedclass · Accepted Answer

$\frac{2 \pi}{5}$

Vedclass · Answer

(A) Since $f(x)$ is continuous at $x = 0$,the limit of the function as $x 	o 0$ must equal the value of the function at $x = 0$. $\lim_{x 	o 0} f(x) = f(0)$ $\lim_{x 	o 0} \frac{4 \sin \pi x}{5 x} = 2k$ We know that $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$. $\lim_{x 	o 0} \left( \frac{4 \sin \pi x}{5 x} ight) = \lim_{x 	o 0} \left( \frac{4 \pi \sin \pi x}{5 \pi x} ight) = \frac{4 \pi}{5} \lim_{x 	o 0} \frac{\sin \pi x}{\pi x} = \frac{4 \pi}{5} 	imes 1 = \frac{4 \pi}{5}$. Equating this to $f(0) = 2k$: $\frac{4 \pi}{5} = 2k$ $k = \frac{4 \pi}{10} = \frac{2 \pi}{5}$ Note: Upon re-evaluating the calculation,the correct value is $\frac{2 \pi}{5}$,which corresponds to option $A$.

If $\begin{aligned} f(x) &= \frac{4 \sin \pi x}{5 x} \text{ for } x \neq 0 \\ &= 2k \text{ for } x = 0 \end{aligned}$ is continuous at $x = 0$,then the value of $k$ is

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