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.

If $\lim_{x\to a}f(x)$ exists (in $\Bbb R$) but is non-zero, and $\lim_{x\to a}g(x)=0$, then
$$
\lim_{x\to a}\frac{f(x)}{g(x)}\quad\text{does not exist.}
$$
Therefore, if $\lim_{x\to 4}\frac{ax-\sqrt{x}+6}{x-4}$ exists, then it must be the case that $\lim_{x\to 4}ax-\sqrt{x}+6=0$. Since the function $x\mapsto ax-\sqrt{x}+6$ is elementary, it is continuous, and so we can evaluate $\lim_{x\to4}ax-\sqrt{x}+6$ by plugging in $x=4$. We find that $\lim_{x\to4}ax-\sqrt{x}+6=4a-\sqrt{4}+6=4a+4$. Hence, $a=-1$.
However, if $a=-1$, then the limit actually equals $-5/4$. To prove this, note that for $x\ge0,x\neq4$, we have
$$
\frac{-x-\sqrt{x}+6}{x-4}=-\frac{x+\sqrt{x}-6}{x-4}=-\frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}=-\frac{\sqrt{x}+3}{\sqrt{x}+2} \, .
$$
Therefore, there is no value of $a$ for which the limit equals $3/4$.
