Question

given the piece - wise function for $f(x)$ above. for which value of “$a$” will $f$ be continuous?
$2$
$4$
$\frac{1}{2}$
$0$
clear my selection
$f(x)=\

$$\begin{cases}a(x - 2)^2&xleq0\\\\sin(2x)&x > 0\\end{cases}$$

$

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = 0$, $\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}a(x - 2)^{2}=a(0 - 2)^{2}=4a$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}\sin(2x)$. Using the fact that $\lim_{u
ightarrow0}\frac{\sin u}{u}=1$, let $u = 2x$. As $x
ightarrow0$, $u
ightarrow0$. So $\lim_{x
ightarrow0^{+}}\sin(2x)=0$.

Step4: Set left - hand and right - hand limits equal

Set $4a=0$, then $a = 0$.

Answer:

$0$