Question
Find the value of $k$ so that the function $f$ is continuous at the indicated point:
$f(x)=\left\{\begin{array}{c}k x^2, \text { if } x \leq 2 \\ 3, \text { if } x>2\end{array}\right.$ at $x=2$.

Answer

Here, $\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} k{x^2} = k \times {2^2}=4k$
Since f(x) is continuous at x = 2 
Therefore, $\mathop {\lim }\limits_{x \to {2^ - }} f(x) = f\left( 2 \right)$
$\Rightarrow 4k = 3$
$ \Rightarrow k = \frac{3}{4}$

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