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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?

Blue
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Jisbon
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  • When you multiply by a conjugate, you want to do it to clear out the "complicated" expression into something that's easier to work with. In this case, it's the $\sqrt x-5$ whose conjugate would make the fraction simpler, even though it is in the numerator. –  Sep 20 '19 at 08:38

4 Answers4

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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 $$\dfrac1{0+10a}$$

lab bhattacharjee
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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

mathsdiscussion.com
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\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}

Andrew Chin
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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 $b=-100$.

Paramanand Singh
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