Find $\lim_{x\to 0} \frac{\sqrt{ax+b}-1}{x}=1$ My answer
Let $\sqrt{ax+b}=y$
Then
$$\lim_{y\to \sqrt b} \frac{(y-1)a}{y^2-b}$$
Let $b=1$
Then $$\lim _{y\to 1} \frac{a}{\frac{y^2-1}{y-1}}$$
$$=\frac a2 =1$$
$$a=2$$
The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the correct answer for this?
 A: $$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}.$$
Now, we see that we need $b=1$ (otherwise, the limit does not exist) and $\frac{a}{\sqrt{b}+1}=1,$
which gives also $a=2$.
If $b\neq1$ for $x\rightarrow0$ we obtain:
$$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}=\frac{a}{\sqrt{ax+b}+1}+\frac{b-1}{x(\sqrt{ax+b}+1)}.$$
We see that for $b>0$ $$\frac{a}{\sqrt{ax+b}+1}\rightarrow\frac{a}{\sqrt{b}+1},$$ but
$$
\lim_{x\rightarrow0}\frac{b-1}{x(\sqrt{ax+b}+1)}$$ does not exist.
A: We can solve this by using series expansion for $\sqrt{ax+b}$. By expanding it, we have
$$
\sqrt{ax +b}= \sqrt{b} +\frac{ax}{2\sqrt{b}} - \frac{a^2x^2}{8b^{3/2}}+ {\large O} $$
(${\large O}$ means other higher powers of $x$ terms).
$$
\lim_{x\to 0} \frac{\sqrt{ax+b} -1}{x}= \frac{\sqrt{b}-1}{x} + \frac{a}{2\sqrt{b}}$$
Now, you can see that for limit to exist we have to have $\sqrt{b}=1 \implies b=1$. And by doing that we find
$$
1= a/2 \\
a=2$$
Hope it helps!
A: Since
$$\lim_{x\to 0}(\sqrt{ax+b}-1) = \lim_{x\to 0}\left(x\frac{\sqrt{ax+b}-1}{x}\right) =0$$
we get
$$\lim_{x\to 0}\sqrt{ax+b}= 1\Rightarrow b=1$$
The limit itself is now the first derivative of $\sqrt{ax+1}$ at $x=0$:
$$\left(\sqrt{ax+1}\right)'(0) = \left. \frac a{2\sqrt{ax+1}}\right|_{x=0} = \frac a{2}\stackrel{!}{=}1 \Rightarrow a=2$$
