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?

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!
