How can I find a number $a$ such that this limit is 1 I want to find the number $a$ and $b$ such that $\lim_{x\to 0} \frac{\sqrt{ax+b}-2}{x}=1$.

First of all, I know that $b$ has to be 4, because the limit of the numerator has to be zero because the denominator is zero when we take its limit.

My problem is with the number 
$a$. I need help.

Thanks a lot!
Multiplying by the conjugate gives:
$$
\frac{\sqrt{ax+b}-2}{x}=\frac{ax+b-4}{x(\sqrt{ax+b}+2)}.
$$

As you already observed, you need $b=4$, for otherwise the limit is infinite.
Then you can simplify by $x$.

Now the limit is:
$$
\frac{a}{\sqrt{4}+2}=\frac{a}{4}.
$$

So you want $a=4$.
multiply by the conjugate on both sides of the equal sign and you get

$$\frac{ax+b-4}x = \sqrt{ax+b}+2$$
multiply by x and plug 4 into b

$$ax+4-4 = x(\sqrt{ax+4}+2)$$
cross out the x's

$$a = \sqrt{ax+4}+2$$
set $x=0$ because $\lim_{x\to 0}$

$$ a = \sqrt{4}+2$$
    $$a=4$$