How can I find the horizontal asymptotes of the function $y=\frac{x-3}{\sqrt{x^2-9}}\:+2$?

Question

It is not a homework question. I ask this question because it was on my test and I genuinely want to understand. I have a very minimal knowledge on the subject of limits. The question is: How can I find the horizontal asymptotes of the function $y=\frac{x-3}{\sqrt{x^2-9}}\:+2$? Help will be much appreciated :)

Answer 1

What you need to do is to calculate both $$\lim_{x\to -\infty}\frac{x-3}{\sqrt{x^2-9}}+2$$ and $$\lim_{x\to +\infty}\frac{x-3}{\sqrt{x^2-9}}+2.$$

Both include and indeterminate of the kind "$\infty/\infty$", and with this kind of functions the usual is to take out a factor $x^k$ both from numerator and denominator, being $k$ the biggest exponent in each case. This lets you compare (or cancel) both $x^k$-factors while the rest tends to a well defined number.

So you can write $x-3=x\left(1-\tfrac3x\right)$ and $$\sqrt{x^2-9}=\sqrt{x^2\left(1-\tfrac9{x^2}\right)}=\sqrt{x^2}\sqrt{1-\tfrac9{x^2})}=|x|\sqrt{x^2\left(1-\tfrac9{x^2}\right)}.$$ But be careful with that $\sqrt{x^2}=|x|$: you can't just cancel out $x$ with $|x|$. Have in mind when checking out both infinite limits that $|x|=x$ for positive $x$ values, but for negative values it happens that $|x|=-x$.

Now try to get the answer: don't forget that 2 over there and let me give you and advance: both limits have different finite result, so there are actually two horizontal asymptotes.

Answer 2

A horizontal asymptote means that a function goes to a limit as $x$ approaches infinity or minus infinity. So we want to determine: $$\lim_{x\to\infty} \frac{x-3}{\sqrt{x^2-9}}+2= 2+\lim_{x\to\infty} \frac{\sqrt{x-3}}{\sqrt{x+3}} = 2 + 1=3.$$ Intuitively you can see that both square roots are approaching infinity with the same rate (the minus 3 and plus 3 are neglibible) as $x$ goes to infinity. Now the other limit: $$\lim_{x\to-\infty} \frac{x-3}{\sqrt{x^2-9}}+2 = 2 - 1=1.$$ Again the numerator and denominator are approaching respectively $-\infty$ and $\infty$ with the same rate.

Therefore you can now say that $y=3$ and $y=1$ are the horizontal asymptotes.