# Rationalized limit denominator, still undefined (divide by zero), how to solve?

I am trying to solve:

$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}}$$

My first step is to multiply by the conjugate to rationalize the denominator.

$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} \cdot \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{4x+1} + \sqrt{5x-1}}$$

Which gives

$$\lim_{x \to 2}\frac {(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})} {({4x+1}) - ({5x-1})}$$

Simplifying denominator

$$\lim_{x \to 2}\frac {(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})} {2-x}$$

Substituting $2$ for $x$ in the denominator and it's zero

$$\lim_{x \to 2}\frac {(\sqrt{x+2} - \sqrt{3x-2}) \cdot (\sqrt{4x+1} + \sqrt{5x-1})} {2-2}$$

Did I make a calculation error, if so where, and / or did I use the wrong approach, if so what's a working approach?

• Try rationalizing the numerator also before substituting 2 for x – AgentS Sep 2 '14 at 20:34
• Ah, multiply by $\frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{x+2} + \sqrt{3x-2}}$? Okay, doing it now, will report back. – antgel Sep 2 '14 at 20:37
• You should have $(\sqrt {4x+1}+\sqrt {5x-1})$ in the numerator. You will see that $\sqrt {x+2}-\sqrt {3x-2}$ becomes $\sqrt 4 - \sqrt 4=0$ and to deal with this you should "rationalise" the numerator too. – Mark Bennet Sep 2 '14 at 20:37
• @topper It's the same multiplication in a different order. Just rationalize what you started with and cancel what needs to be cancelled. Don't bother trying to multiply out the rest. – Mike Sep 2 '14 at 21:02
• \begin{align}\\ &\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}} \times \frac{\sqrt{x+2} + \sqrt{3x-2}}{\sqrt{4x+1} + \sqrt{5x-1}} \times \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \\~\\ &= \frac{(x+2) - (3x-2)}{(4x+1)-(5x-1)} \times \frac{\sqrt{4x+1} + \sqrt{5x-1}}{\sqrt{x+2} + \sqrt{3x-2}} \end{align} – AgentS Sep 2 '14 at 21:39

Since we have \begin{align}\sqrt{x+2}-\sqrt{3x-2}&=\frac{(\sqrt{x+2}-\sqrt{3x-2})(\sqrt{x+2}+\sqrt{3x-2})}{\sqrt{x+2}+\sqrt{3x-2}}\\&=\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\end{align} \begin{align}\sqrt{4x+1}-\sqrt{5x-1}&=\frac{(\sqrt{4x+1}-\sqrt{5x-1})(\sqrt{4x+1}+\sqrt{5x-1})}{\sqrt{4x+1}+\sqrt{5x-1}}\\&=\frac{-(x-2)}{\sqrt{4x+1}+\sqrt{5x-1}}\end{align}
We have $$\frac{\sqrt{x+2}-\sqrt{3x+2}}{\sqrt{4x+1}-\sqrt{5x-1}}=\left(\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\right)\div\left(\frac{-(x-2)}{\sqrt{4x+1}+\sqrt{5x-1}}\right)$$$$=\left(\frac{-2\color{red}{(x-2)}}{\sqrt{x+2}+\sqrt{3x-2}}\right)\times\left(\frac{\sqrt{4x+1}+\sqrt{5x-1}}{-\color{red}{(x-2)}}\right)=2\cdot\frac{\sqrt{4x+1}+\sqrt{5x-1}}{\sqrt{x+2}+\sqrt{3x-2}}$$$$\to 2\cdot \frac{\sqrt 9+\sqrt 9}{\sqrt 4+\sqrt 4}=3\ (x\to 2).$$
• @topper: I thought I might be able to get $x-2$ by having the first two in my answer, which will enable us to get the limit by elinimating $x-2$. – mathlove Sep 2 '14 at 21:00