How to find $a$ and $b$ in $\lim\limits_{x→0}\frac{\sqrt{ax+b}-2}{x}=1$?

Question

How to find $a$ and $b$ such that $$\lim_{x→0}\frac{\sqrt{ax+b}-2}{x}=1?$$

By setting the equation equal to one directly I was able to find out that $b = 4$. However, when I try to solve for $a$ I am stuck and left in circles. How would I go about solving this problem?

Answer 1

$$F=\lim_{x\to0}\dfrac{\sqrt{ax+b}-2}x=\lim_{x\to0}\dfrac{ax+b-4}{x(\sqrt{ax+b}+2)}$$

For the existence of the limit, $b-4=0$

$$\implies1=F=\dfrac a{\sqrt4+2}$$

Answer 2

With $f(x) = \sqrt{ax+b}$ and using $b=4$ as you have already found correctly, for $a$ you may use that

• $\lim_{x\to0}\dfrac{\sqrt{ax+4}-2}x=f'(0)=1$ $$f'(x) = \frac{a}{2\sqrt{ax+4}}\Rightarrow f'(0) =\frac{a}{2\sqrt{4}} = \frac{a}{4} \stackrel{!}{=}1 \Rightarrow \boxed{a= 4}$$

Answer 3

Since the denominator goes to zero it is necessary that also the numerator goes to zero that is $b=4$ and by definition of derivative for $f(x)=\sqrt{ax+4}$ we have

$$\lim_{x\to0}\frac{\sqrt{ax+4}-2}x=f'(0)=\frac a 4=1$$

Answer 4

You could also do it using binomial expansion or Taylor series (built at $x=0$) $$\sqrt{a x+b}=\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right)$$ making $$\frac{\sqrt{ax+b}-2}{x}=\frac{-2+\sqrt{b}+\frac{a x}{2 \sqrt{b}}-\frac{a^2 x^2}{8 b^{3/2}}+O\left(x^3\right) }x=\frac{\sqrt{b}-2}{x}+\frac{a}{2 \sqrt{b}}-\frac{a^2 x}{8 b^{3/2}}+O\left(x^2\right)$$ So $$\frac{a}{2 \sqrt{b}}=1 \qquad \text{and}\qquad \sqrt{b}-2=0$$ then $a,b$.

Answer 5

First observe that we must have $b>0$ so ensure that $ax+b>0$ as $x\to 0$.

Next we have $$\sqrt{b} =\lim_{x\to 0}\sqrt{ax+b}=\lim_{x\to 0}x\cdot\frac{\sqrt{ax+b}-2}{x}+2=0\cdot 1+2=2$$ and hence $b=4$.

Next note that $$a=\lim_{x\to 0}\frac{ax}{x}=\lim_{x\to 0}\frac{ax+b-4}{x}=\lim_{x\to 0}\frac{\sqrt{ax+b} - 2}{x}\cdot(\sqrt{ax+b}+2)=1\cdot(\sqrt{b}+2)=4$$