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!
 A: 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 $a=4$.
A: As $x\rightarrow 0$, the square root behaves as $$\sqrt{ax+b}=\sqrt{b}\sqrt{1+\frac{a}{b}x} \sim \sqrt{b}\left(1+ \frac{1}{2}\frac{a}{b}x+O(x^2)\right).$$
Subtracting the $2$ and dividing by $x$, we have
$$
\frac{\sqrt{ax+b}-2}{x}\sim\frac{\sqrt{b}-2}{x}+\frac{a}{2\sqrt{b}}+O(x).
$$
For the first (diverging) term to vanish, you need $b=4$; to make the second (constant) term equal to $1$, you also need $a=4$.
A: 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
$$\lim_{x \to 0}\left(\frac{\sqrt{4x+4}-2}{x}\right) = 1 \, . $$
A: As you have figured out the value of $ b $ to be $4 $ hence now the limit will take the form of $ 0/0$ . Hence by applying L'Hopitals rule we get
$$ \lim_{x\to 0 } \frac {a}{2\sqrt{ax+4}} = 1$$
Now we can apply limits by direct substitution and after doing that we get 
$$ \frac {a}{4} = 1 $$ 
Solving this we will get $$ a = 4 $$ 
A: 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 $\lim_{x\to 0}$
$$ a = \sqrt{4}+2$$
    $$a=4$$
