Question

Determine the value of the constant k so that the function

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

 

Answer 1

\[f\left( x \right) = \binom{\frac{\sin2x}{5x}, if x \neq 0}{k, if x = 0}\]

If  

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

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

\[\lim_{x \to 0} \frac{\sin2x}{5x} = k\]

\[\lim_{x \to 0} \frac{2\sin2x}{5 \times 2x} = k\]

\[\frac{2}{5} \lim_{x \to 0} \frac{\sin2x}{2x} = k\]

\[\frac{2}{5} \times 1 = k\]

\[k = \frac{2}{5}\]