# Asymptotes

An **asymptote** of a curve is a line to which the curve converges. In other words, the curve and its asymptote get infinitely close, but they never meet. Asymptotes have a variety of applications: they are used in big O notation, they are simple approximations to complex equations, and they are useful for graphing rational equations. In this wiki, we will see how to determine the asymptotes of any given curve.

In most cases, the asymptote(s) of a curve can be found by taking the limit of a value where the function is not defined. Typical examples would be \(\infty\) and \(-\infty,\) or the point where the denominator of a rational function equals zero. Asymptotes are generally straight lines, unless mentioned otherwise. Asymptotes can be broadly classified into three categories: horizontal, vertical and oblique. We will now understand when each type of asymptote occurs.

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## Vertical Asymptotes

Main article: Vertical Asymptotes.

One of the easiest examples of a curve with asymptotes would be \(y=\frac{1}{x}.\) Note that this is a rational function. In order to find its asymptotes, we take the limits of all the values where the function is not defined, which are \(-\infty, 0,\) and \(\infty.\) For \(x\rightarrow 0,\) we should check both the right- and left-hand limits. Then we have

\[\begin{align} \lim_{x\rightarrow-\infty}\frac{1}{x}&=0^{-}\\ \lim_{x\rightarrow\infty}\frac{1}{x}&=0^{+}\\ \lim_{x\rightarrow0^{-}}\frac{1}{x}&=-\infty\\ \lim_{x\rightarrow0^{+}}\frac{1}{x}&=\infty. \end{align}\]

Therefore we can conclude as follows:

(1) When \(x\rightarrow-\infty,\) the curve converges to the lower and negative side of the \(x\)-axis.

(2) When \(x\rightarrow\infty,\) the curve converges to the upper and positive side of the \(x\)-axis.

(3) When \(x\rightarrow0^{-},\) the curve goes infinitely downward, converging to the negative side of the \(y\)-axis.

(4) When \(x\rightarrow0^{+},\) the curve goes infinitely upward, converging to the positive side of the \(y\)-axis.

Thus, the asymptotes of \(y=\frac{1}{x}\) are the \(x\)- and \(y\)-axes. Now take a look for yourself at the graph of the curve, and see if our conclusion corresponds with the figure.

You've probably noticed that simply drawing the graph is the best way to find asymptotes, and you're right! So, if the graph of the curve we are dealing with is familiar, just draw the graph and try to get a hang of where the asymptotes would be. For complex functions, estimating its graph by differentiation would be helpful. Let's take a look at some cases and see how it's done.

## Horizontal Asymptotes

Shown below are the graphs of four different functions, and we can see that the first graph does not have a horizontal asymptote, the second and the third graphs have a horizontal asymptote, and the fourth graph has two horizontal asymptotes. Can you recognize which factors affect the number of horizontal asymptotes of a function?

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- In the first graph, both limits \( \displaystyle \lim _{x \to \infty } f(x) \) and \( \displaystyle \lim _{x \to - \infty } f(x) \) are not finite. This is why there are no horizontal asymptotes in the first graph.
- In the second graph, only one of the limits is finite, and therefore it has only one horizontal asymptote.
- In the third graph, both limits are constant, but both limits are equal, so there is only one horizontal asymptote.
- In the fourth graph, both limits are finite, and both limits are different, so it has two distinct horizontal asymptotes.

Can you think of a function which has three or more horizontal asymptotes?

It is not possible to have a function with more than two horizontal asymptotes. However, we can have curves, which are not functions to have more than two asymptotes.

## Find the horizontal asymptotes of \( y = \arctan x \).

The horizontal asymptotes of \( y = \arctan x \) are \( y = \frac{ \pi } { 2} \) and \( y = - \frac{ \pi } { 2 } \). \(_\square \)

## Oblique Asymptotes

- Examples

## Curvilinear Asymptotes

- Examples

## Example Problems

## What is the asymptote of the curve \(y=(1-x)e^x?\)

Let \(y=f(x),\) then \(f(0)=1\) and \(f(1)=0,\) which gives two points this curve passes through: \((0, 1)\) and \((1, 0).\) The table to be obtained below will also give us some idea of what the curve looks like.

The first and second derivatives of \(f(x)\) are

\[\begin{align} f'(x)&=-e^x+(1-x)e^x=-xe^x \\ f''(x)&=-e^x-xe^x=-(1+x)e^x. \end{align}\]

Letting \(f'(x)=-xe^x=0,\) we have \(x=0.\)

Also, letting \(f''(x)=-(1+x)e^x=0\) gives \(x=-1.\)

Then checking the signs of \(f'(x)\) and \(f''(x)\) around \(x=-1\) and \(x=0,\) we get the following table:\[ \begin{array} { c c c c c c } x & \cdots & -1 & \cdots & 0 & \cdots \\ \\ f'(x) & (+) & (+) & (+) & 0 & (-) \\ f''(x) & (+) & 0 & (-) &(-) & (-) \\ f(x) & \text{(concave up)} & \frac{2}{e} & \text{(concave down)} & 1 & \text{(concave down)} \end{array} \]

Now, checking the limits of \(f(x)\) as our final step, we have

\[\lim_{x \rightarrow \infty} f(x) =-\infty, \lim_{x \rightarrow -\infty} f(x) =0.\]

Thus, in the interval \([-1, \infty),\) the curve is concave down, peaking at \(x=0\) and going down indefinitely as the value of \(x\) approaches \(\infty.\)

In the interval \((-\infty, -1],\) the curve is concave up, getting infinitely close to the \(x\)-axis as the value of \(x\) approaches \(-\infty.\) Using this information, we can roughly draw the graph of \(y=f(x),\) which would look like the figure below:Therefore, the asymptote of the given curve is the \(x\)-axis. \( _\square \)

What are the asymptotes of the graph \(y=\frac{x^2+1}{2x}?\)

Let \(y=f(x),\) then the domain of \(f(x)\) is all the real numbers \(x\) such that \(x\ne 0\) since the denominator cannot be zero. Also, the graph \(y=f(x)\) is symmetrical with respect to the origin since \(f(-x)=-f(x).\)

Now, to have a better idea of what the graph looks like, we get the first and second derivatives of \(f(x)\) as follows:

\[\begin{align} f'(x) &=\frac{(2x) \cdot (2x)-(x^2+1)\cdot 2}{(2x)^2} \\ &=\frac{2x^2-2}{4x^2} \\ &=\frac{(x+1)(x-1)}{2x^2} \\\\ f''(x) &=\frac{(2x)\cdot(2x^2)-(x^2-1)\cdot (4x)}{\left(2x^2\right)^2}\\ &=\frac{1}{x^3}. \end{align}\]

Letting \(f'(x)=\frac{(x+1)(x-1)}{2x^2}=0 ,\) we have \(x=-1\) or \(x=1.\)

Letting \(f''(x)=\frac{1}{x^3}=0\) gives no solution, implying that the graph has no inflection points.Then checking the signs of \(f'(x)\) and \(f''(x)\) around \(x=-1,x=0,\) and \(x=1,\) we get the following table:

\[ \begin{array} { c c c c c c c } x & \cdots & -1 & \cdots &0 & \cdots & 1 & \cdots \\ \\ f'(x) & (+) & 0 &(-) & & (-) & 0 & (+) \\ f''(x) & (-) & (-) &(-) & & (+) &(+) & (+) \\ f(x) &\text{(concave down)} &-1 &\text{(concave down)} & &\text{(concave up)} & 1 & \text{(concave up)} \end{array} \]

Now, observe that \(f(x)\) can be rewritten as \(f(x)=\frac{x^2+1}{2x}=\frac{1}{2x}+\frac{1}{2}x.\) Then checking the limits of \(f(x)\) as our final step gives

\[\lim_{x \rightarrow +0} f(x) =\infty, \lim_{x \rightarrow -0} f(x) =-\infty\]

and

\[\lim_{x \rightarrow \infty} \left(f(x)-\frac{1}{2}x\right)=0, \lim_{x \rightarrow -\infty} \left(f(x)-\frac{1}{2}x\right)=0.\]

Thus, in the interval \((0, \infty),\) the curve is concave down, getting infinitely close to the \(y\)-axis as \(x\rightarrow 0\) and to the line \(y=\frac{1}{2}x\) as \(x\rightarrow \infty.\)

Also, in the interval \((-\infty, 0),\) the curve is concave up, getting infinitely close to the \(y\)-axis as \(x\rightarrow 0\) and to the line \(y=\frac{1}{2}x\) as \(x\rightarrow -\infty.\) Now we can draw the graph of \(y=f(x),\) which would look like the figure below:

Therefore, the asymptotes of the given graph are the \(y\)-axis and the line \(y=\frac{1}{2}x.\) \( _\square \)

What is the asymptote of the curve \(y=2x^2-\ln x?\)

Let \(y=f(x),\) then since logarithmic functions are defined over positive numbers, the domain of \(f(x)\) is all the real numbers \(x\) such that \(x>0.\)

Now, to have a better idea of what the curve looks like, we get the first and second derivatives of \(f(x)\) as follows:

\[\begin{align} f'(x) &=4x-\frac{1}{x}\\&=\frac{4x^2-1}{x}\\&=\frac{(2x+1)(2x-1)}{x} \\\\ f''(x) &=4+\frac{1}{x^2}. \end{align}\]

Letting \(f'(x)=\frac{(2x+1)(2x-1)}{x}=0,\) we have \(x=\frac{1}{2}\) since \(x>0.\)

Letting \(f''(x)=4+\frac{1}{x^2}=0\) gives no solution because \(4+\frac{1}{x^2}>0,\) implying that the curve has no inflection points.Then checking the signs of \(f'(x)\) and \(f''(x)\) around \(x=\frac{1}{2},\) we get the following table:

\[ \begin{array} { c c c c } x &0 & \cdots & \frac{1}{2} & \cdots \\ \\ f'(x) & & (-) & 0 & (+) \\ f''(x) & & (+) & (+) & (+) \\ f(x) & & \text{(concave up)} & \left(\frac{1}{2}-\ln \frac{1}{2}\right) & \text{(concave up)} \end{array} \]

Now, checking the limits of \(f(x)\) as our final step, we have

\[\lim_{x \rightarrow +0} f(x) =\infty, \lim_{x \rightarrow \infty} f(x) =\infty.\]

Thus, in the interval \((0, \infty),\) the curve is concave up, bottoming at \(x=\frac{1}{2}\) and going up indefinitely as the value of \(x\) approaches either \(0\) or \(\infty.\) Now we can roughly draw the graph of \(y=f(x),\) which would look like the figure below:

Therefore, the asymptote of the given curve is the \(y\)-axis. \( _\square \)