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]

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]

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}?

 

[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?