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.

The denominator tends to $0$ and the numerator tends to $4a+4$. If $4a+4 \neq 0$ then the left hand and right hand limits are not equal and the ratio has no limit. Hence we must have $a=-1$.
If $4a+4 >0$ then the right hand limit is $\infty$ and the left hand limit is $-\infty$. If $4a+4 <0$ then the right hand limit is $-\infty$ and the left hand limit is $\infty$.
When $a=-1$ we can apply L'Hopital's Rule to show that the limit is actually $-\frac  5 4$.
