Not getting the right answer in this limit with absolute value

Question

$\lim_{x \to a} \dfrac{\sqrt{ax}-|a|}{ax-a^2}$ , a<0

im getting:

$\lim_{x \to a} \dfrac{1}{\sqrt{ax}+a}$ So my final answer is: $\dfrac{1}{|a|+a}$

But the right answer is: $\dfrac{1}{2|a|}$

Im not sure why, can you help me please?

Answer 1

It should be $$ \lim_{x \to a} \dfrac{\sqrt{ax}-|a|}{ax-a^2}= \lim_{x \to a} \dfrac{\sqrt{ax}-|a|}{ax-a^2}\cdot\dfrac{\sqrt{ax}+|a|}{\sqrt{ax}+|a|}=\lim_{x \to a} \dfrac{1}{\sqrt{ax}+|a|}=\frac{1}{2|a|} $$

Answer 2

$$\frac{\sqrt{ax}-|a|}{ax-a^2} = \frac{\sqrt{ax}-|a|}{(\sqrt{ax}-|a|)(\sqrt{ax}+|a|)} = \frac{1}{\sqrt{ax}+|a|}$$ Now take the limit as $x\rightarrow a$. The above expression approaches $\displaystyle\frac{1}{2|a|}$.