For what value of $k$ is the following function continuous at $x = -\frac{\pi}{6}$?
$f(x) = \begin{cases} \frac{\sqrt{3} \sin x + \cos x}{x + \frac{\pi}{6}}, & x \neq -\frac{\pi}{6} \\ k, & x = -\frac{\pi}{6} \end{cases}$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 ##
$\lim\limits_{x \to -\frac{\pi}{6}} f(x) = \lim\limits_{x \to -\frac{\pi}{6}} \frac{\sqrt{3} \sin x + \cos x}{x + \frac{\pi}{6}}$
$\lim\limits_{x \to -\frac{\pi}{6}} f(x) = \lim\limits_{x \to -\frac{\pi}{6}} \frac{2 \left( \frac{\sqrt{3}}{2} \sin x + \cos x \cdot \frac{1}{2} \right)}{x + \frac{\pi}{6}}$
$\lim\limits_{x \to -\frac{\pi}{6}} f(x) = \lim\limits_{x \to -\frac{\pi}{6}} \frac{2 \sin \left( x + \frac{\pi}{6} \right)}{x + \frac{\pi}{6}}$
$= 2$
$f\left( -\frac{\pi}{6} \right) = k$
For the continuity of $f(x)$ at $x = -\frac{\pi}{6}$,
$f\left( -\frac{\pi}{6} \right) = \lim\limits_{x \to -\frac{\pi}{6}} f(x)$
or $k = 2$