Calculus Midterm Question. I'm studying for my calc midterm right now, and I was finding the limit of the question:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}}\tag{1}$$
When I was trying to solve it I came up with this:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2}^{2} - \sqrt{2x}^{2}}{x-2(\sqrt{x-2}+ \sqrt{2x})}}\tag{2}$$
$$\lim_{x \rightarrow 2}{\frac{x + 2 - 2x}{x-2(\sqrt{x+2}+ \sqrt{2x})}}\tag{3}$$
$$\lim_{x \rightarrow 2}{\frac{-2x}{\sqrt{x+2} + \sqrt{2x}}}\tag{4}$$
However, the solution is this: Solution
I'm not understanding what happens to that $-2x.$
 A: Your error is your move from line $(3)$ to line $(4)$. Those two expressions are not equal.
Multiplying the numerator and denominator by the conjugate of $\;\displaystyle \frac{\sqrt{x+2} - \sqrt{2x}}{ x-2}\;$ gives us:
$$
\begin{align} 
\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}\cdot \frac{\sqrt{x+2} + \sqrt{2x}}{\sqrt{x+2} + \sqrt{2x}}
& = \frac{x+2 - 2x}{(x- 2)(\sqrt{x+2} + \sqrt{2x)}} \tag{a}\\ \\
& = \frac{-(x-2)}{(x- 2)(\sqrt{x+2} + \sqrt{2x)}} \tag{b} \\ \\
& = \frac{-1}{(\sqrt{x+2} + \sqrt{2x)}}\tag{c}\\ \\ 
\end{align}
$$
Now evaluate your $\displaystyle \;\;\lim_{x\to 2} \; \frac{-1}{\sqrt{x+2} + \sqrt{2x}}$.

Note that the $x + 2 - 2x = -x + 2 = -(x-2)$, and so we are able to cancel the factor $(x - 2)$ from both numerator and denominator.
$$\lim_{x\to 2} \frac{\sqrt{x+2} - \sqrt{2x}}{x-2} = \lim_{x\to 2}\;\frac{-1}{(\sqrt{x+2} + \sqrt{2x)}} = \frac{-1}{4}$$
A: L' Hospital's rule can only be applied to a term which goes to $\frac{0}{0}$ (or $\frac{\infty}{\infty}$, which is not applicable here.) Here, the term,
$\frac{x + 2 - 2x}{x-2}$, goes to $\frac{0}{0}$ as $x$ tends to $2$. So the rule applies to this term. You are making a mistake by applying the rule to $\frac{x+2}{x-2}$. Here only the denominator goes to zero, as $x$ tends to $2$, not the numerator.
In any case, I don't see, how you can split up the function in that manner.
