Infinite Limits and Vertical Asymptotes

Key Questions

• The vertical asymptote is a place where the function is undefined and the limit of the function does not exist.

This is because as $1$ approaches the asymptote, even small shifts in the $x$-value lead to arbitrarily large fluctuations in the value of the function.

On the graph of a function $f \left(x\right)$, a vertical asymptote occurs at a point $P = \left({x}_{0} , {y}_{0}\right)$ if the limit of the function approaches $\infty$ or $- \infty$ as $x \to {x}_{0}$.

For a more rigorous definition, James Stewart's Calculus, ${6}^{t h}$ edition, gives us the following:

"Definition: The line x=a is called a vertical asymptote of the curve $y = f \left(x\right)$ if at least one of the following statements is true:

${\lim}_{x \to a} f \left(x\right) = \infty$
${\lim}_{x \to a} f \left(x\right) = - \infty$
${\lim}_{x \to {a}^{+}} f \left(x\right) = \infty$
${\lim}_{x \to {a}^{+}} f \left(x\right) = - \infty$
${\lim}_{x \to {a}^{-}} f \left(x\right) = \infty$
${\lim}_{x \to {a}^{-}} f \left(x\right) = - \infty$"

In the above definition, the superscript + denotes the right-hand limit of $f \left(x\right)$ as $x \to a$, and the superscript denotes the left-hand limit.

Regarding other aspects of calculus, in general, one cannot differentiate a function at its vertical asymptote (even if the function may be differentiable over a smaller domain), nor can one integrate at this vertical asymptote, because the function is not continuous there.

As an example, consider the function $f \left(x\right) = \frac{1}{x}$.

As we approach $x = 0$ from the left or the right, $f \left(x\right)$ becomes arbitrarily negative or arbitrarily positive respectively.

In this case, two of our statements from the definition are true: specifically, the third and the sixth. Therefore, we say that:

$f \left(x\right) = \frac{1}{x}$ has a vertical asymptote at $x = 0$.

See image below. Sources:
Stewart, James. Calculus. ${6}^{t h}$ ed. Belmont: Thomson Higher Education, 2008. Print.

• The vertical asymptote of $y = \frac{1}{x + 3}$ will occur when the denominator is equal to 0. In this case, that will occur at -3, so the vertical asymptote occurs at $x = - 3$. There is no y-coordinate to be included.

For a more thorough explanation behind vertical asymptotes, see here: http://socratic.org/questions/what-is-a-vertical-asymptote-in-calculus? In summary however, vertical asymptotes occur at $x$-values where the limit of the function, either overall or from the right or the left, approaches $\pm \infty$.

An infinite limit is what a functions y value approaches as it approaches infinity or negative infinity

Explanation:

An infinite limit is what a functions y value approaches as the x value approaches infinity or negative infinity

For example
$\lim x \to \infty {e}^{x} = \infty$
$\lim x \to - \infty {e}^{x} = 0$