<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] $$\frac{\sqrt{ax+b}-1}{x}=\frac{ax+b-1}{x(\sqrt{ax+b}+1)}.$$
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.
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