# Math Help - Ugh...Algebra is harder than Calculus

1. ## Ugh...Algebra is harder than Calculus

Need help with some algebraic simplification. This is an example from my calculus book asking to find the vertical and horizontal asymptotes. They divided the numerator and denominator by $x$. I don't understand how they they got to the second step after dividing by x and simplifying. Can someone explain this? Radicals screw everything up for me.

$\lim_{x\to\infty}\frac{\sqrt{2x^2+1}}{3x-5} =$
$\lim_{x\to\infty}\frac{\sqrt{2+(1/x^2)}}{3-(5/x)}$

2. I'll do what I can.

First, you know things like that $\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}$, right? I hope so because I will use little rules like that in this demonstration.

$\lim_{x\to\infty} \frac{\sqrt{2x^2+1}}{3x-5}$.

Dividing the top and bottom by $x$ is like multiplying the top and bottom by $\frac{1}{x}$.

$\lim_{x\to\infty} \frac{\left(\frac{1}{x}\right)\sqrt{2x^2+1}}{\left (\frac{1}{x}\right)(3x-5)}$

Well $\frac{1}{x} = \sqrt{\frac{1}{x^2}}$. This isn't normally completely correct because the right side is always positive while the left could be positive or negative, but we are allowed in this case since we are taking the limit to positive infinity. Therefore at some point we will only care about positive values of $x$ (and extremely high ones, at that) and can ignore cases where this equality isn't necessarily true.

Make ths substitution to put this back into what we're working with:

$\lim_{x\to\infty} \frac{\sqrt{\frac{1}{x^2}} \cdot \sqrt{2x^2+1}}{\left(\frac{1}{x}\right)(3x-5)}$

Now apply the little rule I mentioned at the top (and also distribute the bottom).

$\lim_{x\to\infty} \frac{\sqrt{\left(\frac{1}{x^2}\right)(2x^2+1)}}{3-\frac{5}{x}}$

Distribute inside the radical to get that which was desired:

$\lim_{x\to\infty} \frac{\sqrt{2+(1/x^2)}}{3-(5/x)}$.

3. thank you!