Vertical asymptotes

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We explore functions that “shoot to infinity” near certain points.

Consider the function

\[ f(x) = \frac {1}{(x+1)^2}. \]

While the $\lim _{x\to -1} f(x)$ does not exist, something can still be said.

If $f(x)$ grows arbitrarily large as $x$ approaches $a$, we write

\[ \lim _{x\to a} f(x) = \infty \]

and say that the limit of $f(x)$ is equal to infinity as $x$ goes to $a$.

If $|f(x)|$ grows arbitrarily large as $x$ approaches $a$ and $f(x)$ is negative near $a$, we write

\[ \lim _{x\to a} f(x) = -\infty \]

and say that the limit of $f(x)$ is equal to negative infinity as $x$ goes to $a$.

Which of the following are correct?

$\lim _{x\to -1} \frac {1}{(x+1)^2} = \infty $ $\lim _{x\to -1} \frac {1}{(x+1)^2} \to \infty $ $f(x) = \frac {1}{(x+1)^2}$, so $f(-1) = \infty $ $f(x) = \frac {1}{(x+1)^2}$, so as $x\to -1$, $f(x) \to \infty $

On the other hand, consider the function

\[ f(x) = \frac {1}{(x-1)}. \]

While the two sides of the limit as $x$ approaches $1$ do not agree, we can still consider the one-sided limits. We see $\lim _{x\to 1^+} f(x) = \infty $ and $\lim _{x\to 1^-} f(x) = -\infty $.

If at least one of the following hold:

$\lim _{x\to a} f(x) = \pm \infty $,
$\lim _{x\to a^+} f(x) = \pm \infty $,
$\lim _{x\to a^-} f(x) = \pm \infty $,

then the line $x=a$ is a vertical asymptote of $f$.

A vertical asymptote is given by the equation of a vertical line.

Something to notice from the upcoming examples: To find vertical asymptotes, start with discontinuities. If a function is continuous at a point, the limit will exist. There cannot be a vertical asymptote at such a point. To identify the potential vertical asymptotes, find the discontinuities and points that are not in the domain that are approachable from inside the domain (like endpoints).

Find the vertical asymptotes of

\[ f(x) = \frac {x^2-9x+14}{x^2-5x+6}. \]

Since $f$ is a rational function, it is continuous on its domain. So the only points where the function can possibly have a vertical asymptote are zeroes of the denominator. (These are places the function will not exist, so $f$ cannot be continuous there.)

Start by factoring both the numerator and the denominator:

\[ \frac {x^2-9x+14}{x^2-5x+6} = \frac {(x-2)(x-7)}{(x-2)(x-3)} \]

Using limits, we must investigate what happens with $ f(x)$ when $x\to 2$ and $x\to 3$, since $2$ and $3$ are the only zeroes of the denominator. Write

\begin{align*} \lim _{x\to 2} \frac {\cancel {(x-2)}(x-7)}{\cancel {(x-2)}(x-3)} &= \lim _{x\to 2} \frac {(x-7)}{(x-3)}\\ &= \frac {-5}{-1}\\ &=5. \end{align*}

Now write

\begin{align*} \lim _{x\to 3} \frac {\cancel {(x-2)}(x-7)}{\cancel {(x-2)}(x-3)} &= \lim _{x\to 3} \frac {(x-7)}{(x-3)}\\ &= \lim _{x\to 3}\frac {-4}{x-3}. \end{align*}

Consider the one-sided limits separately.

When $x\to 3^+$, the quantity $(x-3)$ is positive (because $x>3$) and approaches $0$ and the numerator is negative. This is a $\relax \boldsymbol {\tfrac {\#}{0}}$form which is negative for all $x$ sufficiently close to, but greater than, $3$. Therefore, $\displaystyle \lim _{x\to 3^+} f(x) = -\infty $. (Note: Computing this one-sided limit is enough to justify that $x=3$ is a vertical asymptote of the function $f$. We will proceed to calculate the other one-sided limit to see if anything can be said about the two-sided limit.)

When $x\to 3^-$, the quantity $(x-3)$ is negative (because $x<3$) and approaches $0$ and the numerator is negative. This is a $\relax \boldsymbol {\tfrac {\#}{0}}$form which is positive for all $x$ sufficiently close to, but less than, $3$. Therefore, $\displaystyle \lim _{x\to 3^-} f(x) = \infty $. Since the two one-sided limits did not agree, $\displaystyle \lim _{x\to 3}f(x)$ does not exist.

Hence we have a vertical asymptote at $x=3$. Since $\displaystyle \lim _{x\to 2} f(x) = 5$ but $2$ is not in the domain of $f$, the graph of $f$ will have a hole a that point. We know there are no other vertical asymptotes because the function $f$ is continuous on $(-\infty , 2)$, $(2, 3)$, and $(3,\infty )$.

Notice that in this previous example, there was a vertical asymptote at $x=3$, even though the number $3$ was not in the domain of the function $f$. Vertical asymptotes are defined in terms of limits. The important point is that we can approach the number $3$ from within the domain of $f$.

Find the vertical asymptotes of

\[ g(x) = \frac {x^2\sin (x)}{e^{2x}\left (x^2-4x\right )}. \]

This function looks a bit more complicated than in the previous example, but the work required is very similar. We’ll start by factoring the numerator and denominator.

\[ \frac {x^2\sin (x)}{e^{2x}\left (x^2-4x\right )} = \frac {x\cdot x\sin (x)}{e^{2x}x\left (x-4\right )}.\]

The numerator $x^2\sin (x)$ and the denominator $e^{2x}x(x-4)$ are each continuous on $(-\infty , \infty )$. That means the only possibilities to check for vertical asymptotes are those numbers making the denominator zero.

Using limits, we must investigate what happens with $g(x)$ when $x\to 0$ and $x\to 4$, since $0$ and $4$ are the only zeroes of the denominator (since $e^{2x}$ is never zero). We write

\begin{align*} \lim _{x\to 0} \frac {\cancel {x}\cdot x\sin (x)}{\cancel {x}e^{2x}(x-4)} &= \lim _{x\to 0} \frac {x\sin (x)}{e^{2x}(x-4)}\\ &= \frac {0}{1(-4)}\\ &=0. \end{align*}

Since $\displaystyle \lim _{x\to 0}g(x)=0$, there is no vertical asymptote at $x=0$.

We turn our attention to the limit as $x$ approaches $4$. Let us start with the limit from the right.

\[ \lim _{x\to 4^+} \frac {\cancel {x}\cdot x\sin (x)}{\cancel {x}e^{2x}(x-4)} = \lim _{x\to 4^+} \frac {x\sin (x)}{e^{2x}(x-4)} \]

This is a one-sided limit with form $\relax \boldsymbol {\tfrac {\#}{0}}$. We will determine the sign of the fraction for $x$ sufficiently close to, but greater than, $4$.

For the numerator, notice that $\pi < 4 < \dfrac {3\pi }{2}$. That means $\sin (4)$ is the sine value of something in Quadrant III, making it negative. That $x$ is near $4$ makes the $x$ factor in the numerator positive. The numerator is negative for all $x$ sufficiently close to, but greater than, $4$.

For the denominator, $e^{2x}$ is always positive. Since $x>4$, the factor $x-4$ is also positive. The denominator is positive for all relevant values of $x$.

That makes the fraction $\dfrac {x\sin (x)}{e^{2x}(x-4)}$ negative valued for all $x$ sufficiently close to, but greater than, $4$. This means $\displaystyle \lim _{x\to 4^+} g(x) = -\infty $.

There is a vertical asymptote at $x=4$. There are no others, since the only other discontinuity is at $0$ but $\displaystyle \lim _{x\to 0}g(x)=0$.

For our final example, we work with a function whose vertical asymptote is NOT where the denominator is zero.

Find the vertical asymptotes of

\[ f(x) = \dfrac {\ln (x)}{x+5}. \]

The domain of the numerator, $\ln (x)$, is $(0,\infty )$ and it is continuous on that entire interval. The denominator, $x+5$, is continuous on $(-\infty , \infty )$, and is zero only at $-5$. That makes the domain of $f$ to be $(0,\infty )$, and $f$ is continuous on this interval.

There is only one $x$-value where $f$ could have a vertical asymptote: at $x=0$. It is not in the domain of $f$, but is approachable from within the domain. Where the denominator is zero, $x=-5$ is NOT a possibility because $-5$ cannot be approached from within $(0,\infty )$ due to the gap in the number line between $-5$ and $0$.

Since we can only approach $0$ from the right (while staying in the domain of $f$), we should examine the right-hand limit:

\[ \lim _{x\to 0^+} f(x) = \lim _{x\to 0^+} \dfrac {\ln (x)}{x+5}. \]

For the numerator we have $\lim _{x\to 0^+} \ln (x) = -\infty $, while for the denominator we have $\lim _{x\to 0^+} (x+5) = 5$. This means $\lim _{x\to 0^+} f(x) = -\infty $.

There is a vertical asymptote at $x=0$. There are no others because $f$ is continuous on its domain $(0,\infty )$.