Find the horizontal asymptotes of the grpah of the function f defined by
$$ f(x) = \frac{x}{\sqrt{x^2+1}}$$
Solution:
$$ \lim_{x\to +\infty} \frac{\sqrt{x^2}}{\sqrt{x^2 +1}}$$ $$ \lim_{x\to +\infty} \sqrt{\frac{1}{1 + 1/x^2}} = 1$$
Okay, the book says the answer is 1 (I skipped the last step). I'm curious, why did x^2 turn into 1 on the numerator and denominator, and why did the whole fration get a big fraction? I think I know this somewhere in my head, I need refresher!
I added a few more steps to make it more clear $$ \lim_{x\to +\infty} \frac{\sqrt{x^2}}{\sqrt{x^2 +1}}$$ Since $\frac{a^n}{b^n}=\left(\frac ab\right)^n$ $$ \lim_{x\to +\infty} \sqrt{\frac{x^2}{x^2 +1}}$$ Multiply with $1$ $$ \lim_{x\to +\infty} \sqrt{\frac{1/x^2}{1/x^2}\cdot\frac{x^2}{x^2 +1}}$$ Since $\frac ab\cdot\frac cd=\frac{ac}{bd}$ $$ \lim_{x\to +\infty} \sqrt{\frac{1}{1 + 1/x^2}} = 1$$
Since detail has already been given for positive $x$, we deal with negative $x$, which have not been mentioned. If $x$ is negative, then $x=-\sqrt{x^2}$. Then the same argument as the one given shows that $$\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1.$$ So $y=-1$ is also a horizontal asymptote.
Another way of seeing this is that $f(x)$ is an odd function, meaning that $f(-x)=-f(x)$ for all $x$, So the graph of $y=f(x)$ for negative $x$ is obtained by rotating the graph for positive $x$ around the origin through $180^\circ$, or equivalently by reflecting in the origin.
