The value of k for which the function, defined by, $f(x) = \begin{cases} \frac{3x + 4 \tan x}{x} & : x \neq 0 \\ k & : x = 0 \end{cases}$ is continuous at $x = 0$, is

For the function to be continuous at $x = 0$, we need:

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

From the definition, $f(0) = k$.

---

So we need to find:

$\lim_{x \to 0} \dfrac{3x + 4\tan x}{x}$

Splitting the fraction by dividing each term in the numerator by $x$ separately:

$= \lim_{x \to 0} \left[\dfrac{3x}{x} + \dfrac{4\tan x}{x}\right]$

$= \lim_{x \to 0} 3 + \lim_{x \to 0} 4 \cdot \dfrac{\tan x}{x}$

We can split the limit like this because each individual limit exists and is finite.

---

Using the standard result $\lim_{x \to 0} \dfrac{\tan x}{x} = 1$ :

This comes from the fact that $\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$, and since $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$ and $\lim_{x \to 0} \dfrac{1}{\cos x} = 1$, the result follows.

---

Substituting back:

$= 3 + 4 \cdot 1$

$= 3 + 4 = 7$

---

Applying the continuity condition:

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

$7 = k$

$\boxed{k = 7}$