Vertical asymptotes

Functions: Limits and asymptotes

Vertical asymptotes

Now that we have seen what horizontal asymptotes have to do with limits, we will now investigates how we can find vertical asymptotes using limits.

Vertical asymptotes

A function $\blue{f(x)}$ has a vertical asymptote $x=\green a$ if one of the following situations occurs:

$\lim_{x \uparrow \green a}\blue{f(x)}=- \infty$ or $\lim_{x \uparrow \green a}\blue{f(x)}= \infty$

and/or

$\lim_{x \downarrow \green a}\blue{f(x)}=- \infty$ or $\lim_{x \downarrow \green a}\blue{f(x)}= \infty$

Example

$\blue{f(x)}=\blue{\frac{1}{x+2}}$

has a vertical asymptote $x=\green{-2}$ , since

$\lim_{x \uparrow \green{-2}}\blue{\frac{1}{x+2}}=-\infty$ and $\lim_{x \downarrow \green{-2}}\blue{\frac{1}{x+2}}=\infty$

Visual example

In the figure, we see the graph of $\blue{f(x)}=\blue{\frac{1}{x+2}}$ . There is a vertical asymptote at $x=\green{-2}$ because the function approaches the line $x=\green{-2}$ arbitrarily closely, but they will never intersect.

Calculating vertical asymptotes

In order to determine whether a function has a vertical asymptote, we look for values for which the function is not defined. Then we determine the (left and/or right) limit to these values. Useful techniques can be found in The definition of a limit. When the limit satisfies one of the above mentioned situations, we are dealing with a vertical asymptote.

Given the function

$f(x)=\frac{\left(x-2\right)\cdot \left(x+6\right)}{\left(x+2\right)\cdot \left(x+5\right)}$
Determine the vertical asymptotes of this function. Give your answer in the form " $x=b$ " for the right $b$ . If there are multiple vertical asyptotes, select the plus-icon to add an answerfield. If there are no vertical asymptotes, give the answer "none".

$x=-5$ and $x=-2$

The function $f(x)$ has a vertical asymptote in the points where the function value is not defined. In this case this happens when the denominator equals $0$ while the nominator doesn't, namely $x=-5$ and $x=-2$ .

$f(x)=\frac{\left(x-2\right)\cdot \left(x+6\right)}{\left(x+2\right)\cdot \left(x+5\right)}$

In order to verify that these points are vertical asymptotes, we calculate the limit of $f(x)$ where $x$ tends to these values.

$\displaystyle \lim_{x\downarrow \, -5 } f(x) = \displaystyle\lim_{x\downarrow \, -5 } \frac{\left(x-2\right)\cdot \left(x+6\right)}{\left(x+2\right)\cdot \left(x+5\right)}= \infty$

$\displaystyle \lim_{x\downarrow \, -2 } f(x) = \displaystyle\lim_{x\downarrow \, -2 } \frac{\left(x-2\right)\cdot \left(x+6\right)}{\left(x+2\right)\cdot \left(x+5\right)} = -\infty$

We observe that the value of the limits at these points $x=-5$ and $x=-2$ are indeed $\pm\infty$ . Therefore we can conclude that these lines are vertical asymptotes.

In this figure you can see the function and its vertical asymptotes.

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