Asymptotes

What are asymptotes?

In mathematics, an asymptote is a line (or sometimes a curve) that a function’s graph approaches ever more closely as the independent variable x tends toward positive or negative infinity.

In other words, the distance between the graph and the asymptote tends to zero.

• If the asymptote is a straight line, it’s called a linear asymptote.
• If the asymptote is a curve, it’s referred to as a curved asymptote.

There are three types of asymptotes: horizontal, vertical, and oblique (slant).

Horizontal Asymptotes

A horizontal asymptote is determined by evaluating the limit of the function as x approaches positive or negative infinity:

$$ \lim_{x \rightarrow ±∞ } f(x) = c $$

A horizontal asymptote exists if this limit equals a finite real number.

Here’s a simple example of a horizontal asymptote:

In practice, the horizontal asymptote is a line that runs parallel to, or coincides with, the x-axis.

Note that the horizontal asymptotes as x → +∞ and x → −∞ do not necessarily have to be the same.

Example

Let’s determine the horizontal asymptotes of the function:

$$ f(x) = \frac{x+1}{x} $$

First, we compute the limit as x → +∞:

$$ \lim_{x \rightarrow +∞ }\frac{x+1}{x} = 1 $$

Therefore, the function has a horizontal asymptote at y = 1 as x → +∞.

Note. This is an indeterminate form ∞/∞, which can easily be resolved using L'Hôpital’s Rule.

Now, we compute the limit as x → −∞:

$$ \lim_{x \rightarrow -∞ }\frac{x+1}{x} = 1 $$

Thus, the function also has a horizontal asymptote at y = 1 as x → −∞.

 

Vertical Asymptotes

A vertical asymptote occurs at points where the function is undefined. It is determined by evaluating the limit of f(x) as x approaches x₀ from the right and from the left: $$ \lim_{x \rightarrow x_0^+ } f(x) = ±∞ $$ $$ \lim_{x \rightarrow x_0^- } f(x) = ±∞ $$ where x₀ is a point of discontinuity.

A vertical asymptote exists at x₀ if these limits tend to positive or negative infinity.

Here’s a practical example of a vertical asymptote:

In practice, a vertical asymptote is a line parallel to, or coinciding with, the y-axis.

Example

Let’s check whether the following function has any vertical asymptotes:

$$ f(x) = \frac{x^2}{x-1} $$

This function is undefined at x = 1.

We now evaluate the limit as x approaches 1 from both sides:

$$ \lim_{x \rightarrow 1^+ } \frac{x^2}{x-1} = +∞ $$

$$ \lim_{x \rightarrow 1^- } \frac{x^2}{x-1} = -∞ $$

Therefore, the function has a vertical asymptote at x = 1.

Oblique Asymptotes

An oblique (or slant) asymptote exists when the limit of the difference between the function f(x) and the line y = mx + q approaches zero as x tends toward positive or negative infinity: $$ \lim_{x \rightarrow ±∞ } f(x) - (mx+q) = 0 $$

Here’s a practical example of an oblique asymptote:

To determine whether an oblique asymptote exists as x → +∞, we first calculate the slope m:

$$ m = \lim_{x \rightarrow +∞ } \frac{f(x)}{x} \ne 0 $$

Proof. If $$ \lim_{x \rightarrow +∞ } f(x) - (mx+q) = 0 $$ then we can initially set q = 0 and compute: $$ \lim_{x \rightarrow +∞ } f(x) - mx = 0 $$ Since the limit tends to zero as x → ∞, we can safely divide by x: $$ \lim_{x \rightarrow +∞ } \frac{f(x) - mx}{x} = 0 $$ $$ \lim_{x \rightarrow +∞ } \frac{f(x)}{x} - m = 0 $$ $$ \lim_{x \rightarrow +∞ } \frac{f(x)}{x} = m $$

If the slope m exists and is non-zero, we then compute the intercept q:

$$ q= \lim_{x \rightarrow ±∞ } f(x) - mx \ne ∞ $$

Proof. If $$ \lim_{x \rightarrow +∞ } f(x) - (mx+q) = 0 $$ then: $$ \lim_{x \rightarrow +∞ } f(x) - mx - q = 0 $$ $$ \lim_{x \rightarrow +∞ } f(x) - mx = q $$

If q exists and is finite, the function has an oblique asymptote y = mx + q as x → +∞. Otherwise, it does not.

The same method applies to check for an oblique asymptote as x → −∞.

Example

Let’s check whether the following function has an oblique asymptote:

$$ f(x) = \frac{x^2}{x-1} $$

First, we verify that the limit as x → ∞ is infinite:

$$ \lim_{x \rightarrow ∞} \frac{x^2}{x-1} = ∞ $$

Note. This is an indeterminate form ∞/∞, which can easily be resolved using L'Hôpital’s Rule.

Next, we determine whether the slope m is non-zero:

$$ m = \lim_{x \rightarrow ∞} \frac{ \frac{x^2}{x-1} }{x} $$

$$ m = \lim_{x \rightarrow ∞} \frac{x^2}{x(x-1)} $$

$$ m = \lim_{x \rightarrow ∞} \frac{x}{(x-1)} = 1 $$

Since m = 1 is finite and non-zero, we now compute the intercept q:

$$ q = \lim_{x \rightarrow ∞} \frac{x^2}{x-1} - mx $$

Since m = 1:

$$ q = \lim_{x \rightarrow ∞} \frac{x^2}{x-1} - x $$

$$ q = \lim_{x \rightarrow ∞} \frac{x^2-x^2+x}{x-1} $$

$$ q = \lim_{x \rightarrow ∞} \frac{x}{x-1} = 1 $$

Thus, the function has an oblique asymptote with slope m = 1 and intercept q = 1:

And so on.