<s>[INST] 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?
 [/INST] 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 $x=0$:
$$\left(\sqrt{ax+1}\right)'(0) = \left. \frac a{2\sqrt{ax+1}}\right|_{x=0} = \frac a{2}\stackrel{!}{=}1 \Rightarrow a=2$$
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