If f(x) = begin{cases} frac{3 sin(pi x)}{5x} | vedclass

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(A) For a function to be continuous at $x = 0$,the limit of the function as $x 	o 0$ must be equal to the value of the function at $x = 0$. Given $f(

Vedclass · Accepted Answer

$\frac{3\pi}{10}$

Vedclass · Answer

(A) For a function to be continuous at $x = 0$,the limit of the function as $x 	o 0$ must be equal to the value of the function at $x = 0$. Given $f(x) = \frac{3 \sin(\pi x)}{5x}$ for $x 
eq 0$ and $f(0) = 2K$. We know that $\lim_{x 	o 0} \frac{\sin(	heta x)}{x} = 	heta$. Therefore,$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{3 \sin(\pi x)}{5x} = \frac{3}{5} \lim_{x 	o 0} \frac{\sin(\pi x)}{x} = \frac{3}{5} 	imes \pi = \frac{3\pi}{5}$. Since the function is continuous at $x = 0$,we have $\lim_{x 	o 0} f(x) = f(0)$. So,$\frac{3\pi}{5} = 2K$. Solving for $K$,we get $K = \frac{3\pi}{10}$.

If $f(x) = \begin{cases} \frac{3 \sin(\pi x)}{5x} & x \neq 0 \\ 2K & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $K$ is:

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