Find $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$ $\lim\limits_{x \to \infty} 2x(\sqrt{x-1} - \sqrt{x+5})$
For what i've found the part in brackets is an indeterminate form. I've tried to multiply the bracket part by $\frac{\sqrt{x - 1} + \sqrt{x+5}}{\sqrt{x - 1} + \sqrt{x+5}}$ , then multiply the numerator by $2x$. I don't know what to do next.
 A: $$\begin{align}2x (\sqrt{x-1}-\sqrt{x+5}) \cdot \frac{\sqrt{x - 1} + \sqrt{x+5}}{\sqrt{x - 1} + \sqrt{x+5}}& = \dfrac{2x((x-1)-(x+5))}{{\sqrt{x - 1} + \sqrt{x+5}}}\\ \\ &= \dfrac{-12x}{{\sqrt{x - 1} + \sqrt{x+5}}}\tag{1}\end{align}$$
Now you can divide the numerator and denominator by $x$ to get $$\lim_{x\to \infty}\dfrac{-12}{{\sqrt{\frac 1x - \frac 1{x^2}} + \sqrt{\frac 1x +\frac 5{x^2}}}}$$
Alternatively, if it's not clear to you that the limit is $-\infty$, you could instead divide the numerator and denominator $(1)$ by $\sqrt x$ to get $$\begin{align} \lim_{x\to \infty} \dfrac{-12\sqrt x}{\sqrt{1-\frac 1x} + \sqrt{1 +\frac{5}{x}}} & =\lim_{x\to \infty} \dfrac{-12\sqrt x}{2} \\ & = \lim_{x\to \infty} -6\sqrt x = -\infty\end{align}$$
A: You follow the right approach.
\begin{align}
f(x) &= 2x (\sqrt{x-1}-\sqrt{x+5}) = 2x 
\frac{(\sqrt{x-1}-\sqrt{x+5})\cdot (\sqrt{x-1}+\sqrt{x+5})}{(\sqrt{x-1}+\sqrt{x+5})}
\\
&=2x\frac{x-1-x-5}{(\sqrt{x-1}+\sqrt{x+5})} = 2x\frac{-6}{(\sqrt{x-1}+\sqrt{x+5})}
=  -12\frac{x}{(\sqrt{x-1}+\sqrt{x+5})}
\end{align}
Now you can try to bound $f$ from below and above by an expression of the form $$c\frac{x}{\sqrt{x}}$$
What do you know about an expression of that form?
