<s>[INST] Determining $A$ such that $\lim \limits_{x \to\infty }(\sqrt{Ax^2+2x}−5x)$ exists and is finite 
Determine the value of the real number $A$ so that a finite limit exists, and then compute the limit:
$$\lim \limits_{x \to \infty}(\sqrt{Ax^2+2x}−5x)$$

Where would I start with this? I thought I could factor the highest degree variable from each term but that lets any real A work with it.
 [/INST] $\begin{aligned} \sqrt{Ax^2+2x}-5x &= (\sqrt{Ax^2+2x}-5x)\frac{\sqrt{Ax^2+2x}+5x}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{Ax^2+2x-25x^2}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{(A-25)x+2}{\sqrt{A+\frac{2}{x}}+5} \end{aligned}$
Notice that, as $x$ goes to $\infty$, the denominator goes to $\sqrt{A}+5$ and the numerator goes to

*

*$+\infty$ if $A>25$

*$-\infty$ if $A<25$

*$2$ if $A = 25$
so $A=25$ is our only hope to obtain a finite result, which would be
$$\frac{2}{\sqrt{25}+5} = \frac{1}{5}$$
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