Finding $a$ and $b$ such that $\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b} = \frac{1}{40}$ So I am given the following question:

Suppose 
  $$\lim _{x\to 25}\frac{\sqrt{x}-5}{ax+b} = \frac{1}{40}$$ 
  Find $a$ and $b$.

I'm not exactly what to do from here, but what I did was multiplying $$\frac{\sqrt{x}-5}{ax+b}$$ by its conjugate 
($\frac{ax-b}{ax-b}$), resulting in 
$$\frac{100a}{625a^2-b^2} = \frac{1}{40}$$
Now I'm stuck and not sure what to do from here. Am I on a correct track?
 A: Let $\sqrt x-5=y$
$$\lim _{x\to 25}\frac{\left(\sqrt{x}-5\right)}{ax+b} =\lim_{y\to0}\dfrac y{a(5+y)^2+b}=\lim_{y\to0}\dfrac1{\dfrac{25a+b}y+10a+ay}$$
For the existence of the limit $25a+b=0$ as the numerator $\to0$
In that case the limit reduces to $$\dfrac1{0+10a}$$
A: As numerator is zero for limit to exist denominator should be zero at x= 25 . $\\$  therefore b=-25a $\\$
Also now $\lim_{x\to 25}(\frac{\sqrt{x}-5}{a(x-25)}) ={\frac{1}{40}} $. $\\$
Therefore $\lim_{x\to 25}{\frac{1}{a({\sqrt{x}+5})}}=\frac{1}{40} $. $\\$
$10a=40$. $\\$
Hence a=4  and b=-100
A: \begin{align}
\frac1{40}=\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b}&=\lim_{x\to25}\frac{\sqrt{x}-5}{ax+b}\cdot\frac{\sqrt{x}+5}{\sqrt{x}+5}\\
\frac1{40}&=\lim_{x\to25}\frac{x-25}{(ax+b)(\sqrt{x}+5)}
\end{align}
Observe that by substituting $x=25$ into $\sqrt{x}+5$ we get $10$.  Therefore, that missing factor of $4$ that we want (to make the denominator $40$) must come from $ax+b$.  We also know that $x-25$ is a factor of $ax+b$.  Thus, we can write
\begin{align}
ax+b&=4(x-25)\\
ax+b&=4x-100
\end{align}
Therefore, $\boxed{(a,b)=(4,-100)}$.  To show that this is the solution, we can evaluate the limit:
\begin{align}
\lim_{x\to25}\frac{\sqrt{x}-5}{4x-100}&=\lim_{x\to25}\frac{\sqrt{x}-5}{4(x-25)}\cdot\frac{\sqrt{x}+5}{\sqrt{x}+5}\\
&=\lim_{x\to25}\frac{x-25}{4(x-25)(\sqrt{x}+5)}\\
&=\lim_{x\to25}\frac{1}{4(\sqrt{x}+5)}\\
&=\frac{1}{4(\sqrt{25}+5)}\\
&=\frac{1}{4(10)}\\
&=\frac{1}{40}
\end{align}
A: You can proceed using algebra of limits. We have $$25a+b=\lim_{x\to 25}ax+b=\lim_{x\to 25}\dfrac{1}{\dfrac{\sqrt {x}-5}{ax+b}} \cdot(\sqrt {x} - 5)=40\cdot 0=0$$ Therefore we have $b=-25a$ and we are given that $$\lim_{x\to 25}\frac{\sqrt{x} - 5}{a(x-25)}=\frac{1}{40}$$ This ensures that $a\neq 0$ and $$a=40\lim_{x\to 25}\frac{\sqrt{x}-5}{x-25}=4$$ and then $b=-100$.
