# 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!

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@Amzoti $a$ and $b$ need to be numbers, i.e. constants, i.e. independent of $x$. – Fly by Night Jan 30 '13 at 20:22

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 $a=4$.

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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 $a=4$.

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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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Except when $b=4$. – 1015 Jan 30 '13 at 20:12