# Horizontal asymptotes - approaches from above or below?

#### fp252

I seem to be having a lot of difficulty finding whether for a horizontal asymptote, whether the curve approaches the asymptote from above or below.

For example, for the problem $$y = \frac{6x + 1}{1 - 2x}$$, I know that:

For the vertical asymptote, x = 1/2, and that $$\lim_{x \to \frac{1}{2}^{+}} = - \infty$$, while $$\lim_{x \to \frac{1}{2}^{-}} = \infty$$.

Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

This is how I tried to find it:
As x --> -3 from the left, eg. x = -10, $$\frac{6(-3)+1}{1-2(-10)}$$ = $$\frac{-17}{21}$$ = -0.81
therefore approaches from above, because -0.81 is greater than -3.

As x --> -3 from the right, eg. x = 10, $$\frac{6(-3)+1}{1-2(10)}$$ = $$\frac{-17}{-19}$$ = 0.95
therefore also approaches from above, because -0.81 is greater than -3.

However, my graphing calculator says otherwise. The graph shows that, left of the vertical asymptote, the curve approaches from above; to the right, it approaches from the bottom. What am I doing wrong?

#### gabbagabbahey

Homework Helper
Gold Member
Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

This is how I tried to find it:
As x --> -3 from the left, eg. x = -10, $$\frac{6(-3)+1}{1-2(-10)}$$ = $$\frac{-17}{21}$$ = -0.81
therefore approaches from above, because -0.81 is greater than -3.

As x --> -3 from the right, eg. x = 10, $$\frac{6(-3)+1}{1-2(10)}$$ = $$\frac{-17}{-19}$$ = 0.95
therefore also approaches from above, because -0.81 is greater than -3.
Why are you substituting x=-3 into the numerator and then substituting $x=\pm10$ into the denominator?

#### fp252

Hmm, that's what I saw in my class notes. I assumed that the numerator's x-value remained to be -3, while the denominator was supposed to have a "test" value.

Ok, so I'm guessing I'm supposed to be subbing $x=\pm10$ into both numerator and denominator?

#### gabbagabbahey

Homework Helper
Gold Member
Ok, so I'm guessing I'm supposed to be subbing $x=\pm10$ into both numerator and denominator?
Yes! After all, it's y that approaches -3 not x.

#### fp252

Thanks for clarifying, gabbagabbahey!

I don't know how I've been pondering over this simple mistake, haha... #### fp252

Also, another question: is what I shown above involving the calculations of $$\lim_{x \to - \infty}$$ and $$\lim_{x \to +\infty}$$?

Last edited:

#### Gregg

$\displaystyle \lim_{x\to\infty} \frac{6x+1}{1-2x}$

$\displaystyle \lim_{x\to\infty} \frac{6+\frac{1}{x}}{\frac{1}{x}-2}$

Does that help?

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