How to set up equation to find the unknown values if a limit exist.

Question

For what values of the constants $a$ and $b$ does the following limit exist? $$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x$$

for this question, $$f(x) = \frac{-(x+3)(\sqrt{ax+b}-2)}x,x<-3$$ $$f(x) = \frac{(x+3)(\sqrt{ax+b}-2)}x,x>=-3$$

Firstly, I found difficulties in setting up two equation solving for a and b. Secondly, since $|x+3|=x+3$ only when $x\to0$, I can only use the follwoing to set up the equation. $$\lim_{x\to0}\frac{(x+3)(\sqrt{ax+b}-2)}x$$ However, I cannot simplified it to cancel out the denominator $x$ by some method like rationalization. It gets $$\lim_{x\to0}\frac{(x+3)(ax+b-4)}{(\sqrt{ax+b}-2)x}$$. Therefore, how can we set up the equation to find $a$ and $b$ or it needs other methods to do this type of questions?

Answer 1

$\lim_{x\to0}\frac{|x+3|(\sqrt{ax+b}-2)}x $

Since $\lim_{x\to0} |x+3| =3$, that doesn't matter.

What is left is $\lim_{x\to0}\frac{(\sqrt{ax+b}-2)}x $.

For this limit to exist, we must have $\lim_{x\to0}\sqrt{ax+b}-2 =0 $ or $\lim_{x\to0}\sqrt{ax+b} =2 $.

Since $\lim_{x\to0}ax =0 $, we must have $b=4 $.

The answer is, therefore, $b=4$ and any value of $a$.

Answer 2

Note that we can write

$$\begin{align} \frac{|x+3|(\sqrt{ax+b}-2)}{x}&=\left(\frac{|x+3|(\sqrt{ax+b}-2)}{x}\right)\left(\frac{(\sqrt{ax+b}+2)}{(\sqrt{ax+b}+2)}\right)\\\\ &=\frac{|x+3|(ax+b-4)}{x(\sqrt{ax+b}+2)}\\\\ &=\left(\frac{|x+3|}{(\sqrt{ax+b}+2)}\right)\left(a+\frac{b-4}{x}\right)\tag 1 \end{align}$$

The first parenthetical term on the right-hand side of $(1)$ approaches $\frac{3}{2+\sqrt b}$ as $x\to 0$. The limit of the second term does not exist unless $b=4$. If $b=4$, then that limit is $a$, and we find

$$\lim_{x\to 0}\frac{|x+3|(\sqrt{ax+4}-2)}{x}=\frac{3a}{4}$$