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?
Best Answer
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 $$\frac{\sqrt{x+2}-\sqrt{3x+2}}{\sqrt{4x+1}-\sqrt{5x-1}}=\left(\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\right)\div\left(\frac{-(x-2)}{\sqrt{4x+1}+\sqrt{5x-1}}\right)$$$$=\left(\frac{-2\color{red}{(x-2)}}{\sqrt{x+2}+\sqrt{3x-2}}\right)\times\left(\frac{\sqrt{4x+1}+\sqrt{5x-1}}{-\color{red}{(x-2)}}\right)=2\cdot\frac{\sqrt{4x+1}+\sqrt{5x-1}}{\sqrt{x+2}+\sqrt{3x-2}}$$$$\to 2\cdot \frac{\sqrt 9+\sqrt 9}{\sqrt 4+\sqrt 4}=3\ (x\to 2).$$