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

Question

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

Aicharya · Accepted Answer

Since $f(x)$ is continuous at $x=0$, $\begin{array}{l} \lim _{x ightarrow 0} f(x)=f(0) \Rightarrow \lim _{x ightarrow 0} \frac{4 \sin \pi x}{5 x}=2 k \ 	herefore \lim _{x ightarrow 0}\left(\frac{4 \sin \pi x}{\pi x}ight) \cdot \frac{\pi}{5}=2 k \Rightarrow(1) \cdot \frac{4 \pi}{5}=2 k \Rightarrow k=\frac{2 \pi}{5} \end{array}$

If $\begin{aligned} f(x) &=\frac{4 \sin \pi x}{5 x} \text { for } x \neq 0 \\ &=2 \mathrm{k} \quad \text {…

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