Question

In the following, determine the value of constant involved in the definition so that the given function is continuou:  \[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if }  x \neq 0 \\ 3k , & \text{ if  } x = 0\end{cases}\] 

Answer 1

Given:

\[f\left( x \right) = \begin{cases}\frac{\sin 2x}{5x}, & \text{ if }  x \neq 0 \\ 3k , & \text{ if  } x = 0\end{cases}\] 

If  

\[f\left( x \right)\]  is continuous at x = 0, then 

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]

\[\Rightarrow \lim_{x \to 0} \frac{\sin 2x}{5x} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{2\sin 2x}{2 \times 5x} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{5} \lim_{x \to 0} \frac{\sin 2x}{2x} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2}{5} = 3k\]
\[ \Rightarrow k = \frac{2}{15}\]