What is the formal way find $a$ such that $\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$? So I got this problem

Determine the value of $a$ that satisfies the following limit
$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$$

If we substitute $x = 4$, the value of the limit will be $\frac{4a + 4}{0}$ which is undefined. So, the numerator must have the same root as the denominator. With $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$, and the one which make the limit undefined is $(\sqrt{x} - 2)$. From here, I conclude that the numerator must have ($\sqrt{x} - 2$) as a root. And from here, I can do polynomial division and got that $a = -1$.
The other way is by argue that "if the limit have a value, then make the result indeterminate form: $\frac{0}{0}$. Then, we'll get $4a + 4 = 0$ and $a = -1$.
The problem is, I find my method is not good enough for essay problems.

By $y=x-4 \to 0$ we have
$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4}=\lim_{y\to 0} \frac{a(y+4)-\sqrt{y+4}+6}{y}=\lim_{y\to 0} \frac{ay+4a-\sqrt{y+4}+6}{y}$$
and to guarantee that limit exists we need
$$4a-\sqrt{y+4}+6 \to 0 \iff a=-1$$
and for $a=-1$
$$ \frac{-y-\sqrt{y+4}+2}{y}=-1-\frac{\sqrt{y+4}-2}{y} \to -1-\frac14=-\frac 5 4$$
indeed by $f(y)=\sqrt{y+4} \implies f'(y)=\frac1{2\sqrt{y+4}}$ we have that by the definition of derivative
$$\lim_{y\to 0}\frac{\sqrt{y+4}-2}{y}=f'(0)=\frac14$$
