Limits Plus: II

First rationalize the numerator:

$\lim _{ x\rightarrow 0}{\frac{\sqrt{ax+b}-2}{x}\cdot \frac{\sqrt{ax+b}+2}{\sqrt{ax+b}+2}}=\lim _{ x\rightarrow 0 }{\frac{ax+b-4}{x(\sqrt{ax+b}+2)}} .$

Now since the denominator approaches $0$ as $x\rightarrow 0$ , the limit will exist only if the numerator also approaches $0$ as $x\rightarrow 0$ . So we require that

$a(0)+b-4=0 \Rightarrow b=4.$

So the equation becomes

$\lim _{ x\rightarrow 0 }{\frac{a}{\sqrt{ax+b}+2}}=1$

$\Rightarrow \frac{a}{\sqrt{4}+2}=1$

$\Rightarrow a=4.$

Therefore, $a=b=4$ , and,

$a+b=\boxed{8}$ .

$\lim_{x \rightarrow 0}\frac{\sqrt{ax+b}-2}{x}=1$ means that $\sqrt{ax+b}-2$ and $x$ will be closer and closer to each other as x gets closer to 0.

When $\sqrt{ax+b}-2$ and $x$ equal to each other,

$\sqrt{ax+b}-2 =x$

$\sqrt{ax+b} =x+2$

square both side, $ax+b=x^2+4x+4$

$x^2+(4-a)x+(4-b)=0$

As x goes to 0, $\lim_{x \rightarrow 0}x^2+(4-a)x+(4-b)=8-a-b$

As $\sqrt{ax+b}-2$ and $x$ tend to equal to each other, $x^2+(4-a)x+(4-b)$ goes to 0

so, $8-a-b=0$

a+b=8