# Horizontal asymptotes

Thanks.

y=\frac{\left(1-e^x\right)}{\left(1+e^x\right)}

has two horizontal asymptotes at y = -1 AND y=1. I do get that at y=-1 it asymptotes to x->infinity and as x->-infinity it is asymptotes to y = 1.

However, I do not get how to identify if it would be above or below the asymptote. Here, I guessed it but what is the procedure?

Thanks

. . . if it would be above or below the asymptote

Let’s first make sure you don’t have a common misunderstanding of limits, which is the belief that a function can never reach or cross a limiting value. That’s false, for reasons you can read about in our Limits Introduction, and super-important to understand going forward.

That said, you’re right that this function doesn’t cross either asymptote and so we can reason about whether it lies above or below each in the relevant regions.

Here’s how I’d think about it: As x \to -\infty, the e^x terms get smaller and smaller so we end up with essentially 1/1 (which is how you probably reasoned about the limit), but the numerator will always be smaller than the denominator [due to the (1 - small positive number) as opposed to the (1 + small positive value)] and so the fraction will always be less than one. Hence we’re below y = 1 to the left.

For x \to +\infty, it might be easier to view the function as

-\,\frac{(e^x - 1)}{e^x + 1}

Ignoring the overall negative sign for a moment, again the numerator will always be smaller than the denominator, so \frac{(e^x - 1)}{e^x + 1} will be smaller than 1. You can thus imagine that fraction’s value at some point being 0.99999. That makes the overall function’s value -0.99999, which lies above the y = -1 line.

One general tip here can be to substitute a few actual values in to see how the function’s output trends as the input-values increase in both directions, kinda like I just did.

More importantly, you’ll also learn more tools in the coming weeks to help determine the actual shape of a function’s graph. (Good stuff!)

For now, hope that helps!

Hi - thanks for the immediate response.

I got it. Could you please pass me on the exact link where the shape of the function’s graph is discussed? I get a bit lost finding it on the website.

Thanks.
Gowri