Is the answer infinity?

Okay, think of it this way:

We know that $x^2 + 8x + 1 = x^2 + 8x + 16 - 15 = (x+4)^2 - 15$ . When $x$ becomes super duper large (in other words, as $x$ tends to infinity), $(x+4)^2$ becomes super duper ultra awesome large right? And that super duper ultra awesome large minus a small number of 15 is still a super duper ultra awesome large number right? So we say that for $x$ becomes unboundedly large, $(x+4)^2 - 15 \approx (x+4)^2$ .

Now, back to the question, we want to evaluate the limit as $x$ becomes unboundedly large in the positive direction (in other words, as $x$ tends to $\infty$ ), then $\sqrt{x^2+8x+1} = \sqrt{(x+4)^2 - 15}$ is approximately equals to $\sqrt{(x+4)^2}$ right? And recall that $\sqrt{x^2} = x$ for positive $x$ , then we can simplify the expression $\sqrt{(x+4)^2}$ to just $x+4$ instead. But wait, we still need to subtract it by $x$ right? What's left is just $(x+4) - x = 4$ and that's your answer!

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Alternatively, here's another short solution:

Let $y = \dfrac 1x$ , then when $x\to \infty$ , $y\to 0$ , the limit becomes

$\lim_{y\to0} \sqrt{ \dfrac1{y^2} + \dfrac8y + 1} -\dfrac1y= \lim_{y\to0}\dfrac{\sqrt{1+8y+y^2} - 1}y$

What's left to do is to apply L'hopital's rule : Differentiate top and bottom (separately) and we get,

$\lim_{y\to0}\dfrac{2y + 8}{2\sqrt{1+8y+y^2}} =\dfrac{2(0) + 8}{2\sqrt{1+8(0)+0^2}} = 4$

as well!