Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin 5x}{x^2+2x},& x\ne 0\\ k+\frac{1}{2},& x=0\end{array}$ is continuous at $x=0$, then the value of $k$ is

• A. $-2$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

For continuity at $x=0$, we must have $f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{\sin 5x}{x^2+2x}$
$=\underset{x\to 0}{lim}\frac{5x\cdot \frac{\sin 5x}{5x}}{x^2+2x}$
$=\underset{x\to 0}{lim}\frac{5}{x+2}=\frac{5}{2}$
$\Rightarrow k+\frac{1}{2}=\frac{5}{2}$
$\therefore k=2$