How to find $a$ and $b$ given this limit if $f(x)=ax+b$? Question
Suppose $f(x)=ax+b$ and $\lim\limits_{x \to 25} \dfrac{\sqrt{x}-5}{f(x)}=\dfrac{1}{20}$.
Find $a$ and $b$.
Attempt
I figured out that
$\lim\limits_{x \to 25} \sqrt{x}-5=0\space$ but $0$ divided by any denominator will give $0.\quad$ So I do not know how to find $a$ and $b$ given the limit is $\dfrac{1}{20}$.
 A: Since $f(x)=ax+b$ is a polynomial, for all real $c$, we have, $$\lim_{x\to c} f(x)=f(c).$$
If $f(25)\neq0$,
$$\lim_{x\to25}\frac{\sqrt{x}-5}{f(x)}=0.$$
Hence, $f(25)=0$, so,
$$\lim_{x\to25}\frac{\sqrt{x}-5}{f(x)}=\frac{\sqrt{25}-5}{f(25)}=\frac{0}{0}$$
Applying L'Hôpital's rule,
$$\lim_{x\to25}\frac{\sqrt{x}-5}{f(x)}=\lim_{x\to25}\frac{\dfrac{\mathbb{d}}{\mathbb{d}x}\left(\sqrt{x}-5\right)}{f'(x)}=\lim_{x\to 25}\frac{\frac{1}{2\sqrt{x}}}{a}=\frac{1}{20}$$
Therefore, $a=2$. Also, $f(25)=0\implies 25\cdot 2+b=0\implies b=-50. $
A: Following the hints in the comments, $ax+b$ must go to $0$ when $x$ goes to 25. So $$\lim_{x\to 25}ax+b=25a+b=0$$
That way the limt has the form of $0/0$:$$\lim_{x\to 25}\frac{\sqrt x-5}{f(x)}\to\frac 00$$
Since $25$ is a root of $ax+b$, we can write $$f(x)=a(x-25)=a(\sqrt x -5)\sqrt x+5)$$
It is equivalent of saying $b=-25a$.
To find $a$, we calculate the limit:
$$\lim_{x\to 25}\frac{\sqrt x-5}{f(x)}=\lim_{x\to 25}\frac{\sqrt x-5}{a(\sqrt x-5)(\sqrt x+5)}=\frac1{a(\sqrt{25}+5)}=\frac1{10a}=\frac1{20}$$
So $a=2$ and $b=-50$
