If the function $f(x) = \begin{cases}\frac{\sin 3x}{x}, & \text{if } x \neq 0\\ \frac{3k}{2}, & \text{if } x = 0\end{cases}$ is continuous at $x = 0$, then the value of $k$ is

For $f(x)$ to be continuous at $x = 0$:

$\lim_{x \to 0} f(x) = f(0)$

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From the given function:

$f(0) = \frac{3k}{2}$

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When $x \neq 0$:

$f(x) = \frac{\sin 3x}{x}$

Finding the limit:

$\lim_{x \to 0} \frac{\sin 3x}{x}$

$= \lim_{x \to 0} \frac{3 \sin 3x}{3x}$

$= 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x}$

Let $u = 3x$. When $x \to 0$, then $u \to 0$.

$= 3 \cdot \lim_{u \to 0} \frac{\sin u}{u}$

$= 3 \cdot 1$

$= 3$

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Applying the continuity condition:

$\lim_{x \to 0} f(x) = f(0)$

$3 = \frac{3k}{2}$

$6 = 3k$

$k = 2$

Therefore, the value of $k$ is $2$.