Finding the limit of a rational expression

[Mr Davis 97]

To find the horizontal asymptotes of a rational function, we find the limit as x goes to infinity. Given the rational function $\displaystyle\frac{x + 1}{\sqrt{x^2+1}}$, we can find the limit by multiplying the numerator and the denominator by $\frac{1}{x}$. This gives us $\frac{1 + \frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}}$. Taking the limit of this gives us 1, so it would seem as though the horizontal asymptote is 1. However, looking at the original function, it is obvious that if we went in the negative direction to infinity the number would be negative. Therefore, what I am doing wrong? Why doesn't the process of dividing the numerator and denominator yield the correct answer?

 

[PFuser1232]

To find the horizontal asymptotes of a rational function, we find the limit as x goes to infinity. Given the rational function $\displaystyle\frac{x + 1}{\sqrt{x^2+1}}$, we can find the limit by multiplying the numerator and the denominator by $\frac{1}{x}$. This gives us $\frac{1 + \frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}}$. Taking the limit of this gives us 1, so it would seem as though the horizontal asymptote is 1. However, looking at the original function, it is obvious that if we went in the negative direction to infinity the number would be negative. Therefore, what I am doing wrong? Why doesn't the process of dividing the numerator and denominator yield the correct answer?

You can only divide whatever's under the radical sign by $x^2$ if you're sure that $\frac{1}{x}$ is positive, which isn't always the case. Hint: $\sqrt{\frac{1}{x^2}}$ is, in general, equal to $|\frac{1}{x}|$, NOT $\frac{1}{x}$.

 

[Mark44]

In your rational function, the denominator is always positive, but the numerator is positive if x > -1, and is negative if x < -1. There are two horizontal asymptotes: y = 1 (as x → ∞) and y = -1 (as x → -∞).

If you factor the numerator and denominator, you get $\frac{x(1 + 1/x)}{|x|\sqrt{1 + 1/x^2}}$.
$\frac x {|x|} = 1$ if x > 0, but $\frac x {|x|} = -1$ if x < 0.

Note that $\sqrt{x^2} = |x|$, not x.