<s>[INST] 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!
 [/INST] When $a=b=4$. We can calculate the binomial series of $\sqrt{ax+b}$ and we find that
$$\sqrt{ax+b} \sim \sqrt{b} + \frac{a}{2\sqrt{b}}x + \cdots$$
Using this approximation, we see that
$$\frac{\sqrt{ax+b}-2}{x} \sim \frac{\sqrt{b}-2}{x} + \frac{a}{2\sqrt{b}} + \cdots $$
where the tail "$+\cdots$" consists of terms divisible by $x$. Assuming that $\sqrt{b} \neq 2$, the limit is undefined as $x \to 0$. If $\sqrt{b} = 2$, i.e. $b=4$ then we have:
$$\frac{\sqrt{ax+4}-2}{x} \sim \frac{a}{4}-\frac{a^2}{64}x + \cdots $$
where the tail "$+\cdots$" consists of terms divisible by $x^2$. In this case, the limit as $x \to 0$ is $\frac{1}{4}a$ and so we need $a=4$. It follows that
$$\lim_{x \to 0}\left(\frac{\sqrt{4x+4}-2}{x}\right) = 1 \, . $$
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