Explainer: One-Sided Limits

In this explainer, we will learn how to evaluate one-sided limits graphically and algebraically.

Suppose that the function 𝑓 is defined near π‘₯=π‘Ž, but not necessarily at π‘Ž itself.

The easiest way to get such a function is to start with another function, say 𝑔, which is defined everywhere, and then define another function π‘“βˆΆβ„βˆ’{π‘Ž}βŸΆβ„ by setting 𝑓(π‘₯)=𝑔(π‘₯) for every π‘₯β‰ π‘Ž.

Notice that 𝑓 and 𝑔 are different functions, because they do not have the same domain. For example, taking 𝑔(π‘₯)=2π‘₯+1 and π‘Ž=πœ‹, we have the function 𝑓 whose graph is shown in the figure.

The hollow circles at (πœ‹,0) and (πœ‹,2πœ‹+1) indicate that πœ‹ is not in the domain of the function 𝑓. Of course, the graph of 𝑔 is just the entire line with the point (0,2πœ‹+1) included, and 7.283185307179586β€¦β‰ˆ2πœ‹+1.

The idea of β€œthe limit of the values 𝑓(π‘₯) as π‘₯ approaches πœ‹ from the left” is that we are looking for a number 𝐿 so that 𝑓(π‘₯) gets closer and closer to 𝐿 as π‘₯ moves through values in the domain of 𝑓, but only those values that satisfy π‘₯<πœ‹.

For example, we could look at the values π‘₯=3.1,3.14,3.141,3.1415,3.14159,3.141592,3.1415926,3.14159265,… along the π‘₯-axis, which are approaching πœ‹, one decimal at a time, and follow 𝑓(π‘₯)=2π‘₯+1 along the 𝑦-axis, through values 𝑓(π‘₯)=7.2,7.28,7.282,7.283,7.28318,7.283184,7.2831852,7.28318530,….

We see that the limiting value 𝐿 appears to be just 2πœ‹+1. This is written as limο—β†’οŽ„οŽͺ𝑓(π‘₯)=2πœ‹+1 and read as β€œthe one-sided limit of 𝑓(π‘₯) as π‘₯ tends to πœ‹ from the left (or below) is 2πœ‹+1.”

This fact does not require a listing of values as we did above: it is fairly clear from the graph, as is the fact that the limit as we approach πœ‹ from the right (above) is also 2πœ‹+1, which we write as limο—β†’οŽ„οŽ©π‘“(π‘₯)=2πœ‹+1.

A more advancedβ€”and commonβ€”example of a function defined everywhere except at πœ‹ is one given by a piecewise definition: 𝑓(π‘₯)=5.5π‘₯<πœ‹,2π‘₯+1π‘₯>πœ‹,ifif which has the following graph.

Now, just from the graph, the right-sided limit is the same as before, limο—β†’οŽ„οŽ©π‘“(π‘₯)=2πœ‹+1, and the left-sided limit can only be the given constant value limο—β†’οŽ„οŽͺ𝑓(π‘₯)=5.5.

This second case could also have been a piecewise-defined function that was defined even at πœ‹. For example, 𝐴(π‘₯)=5.5π‘₯<πœ‹,2π‘₯+1π‘₯β‰₯πœ‹,𝐡(π‘₯)=5.5π‘₯β‰€πœ‹,2π‘₯+1π‘₯>πœ‹,𝐢(π‘₯)=5.5π‘₯<πœ‹,3π‘₯=πœ‹,2π‘₯+1π‘₯>πœ‹,ififorififorififif all of which have different values 𝐴(πœ‹), 𝐡(πœ‹), and 𝐢(πœ‹). But all have the same behavior as 𝑓 regarding one-sided limits: limlimlimlimο—β†’οŽ„ο—β†’οŽ„ο—β†’οŽ„ο—β†’οŽ„οŽ©οŽ©οŽ©οŽ©π΄(π‘₯)=𝐡(π‘₯)=𝐢(π‘₯)=2πœ‹+1=𝑓(π‘₯), and similarly for the left-sided limits. This is because those limits are only interested in how the functions behave away from π‘₯=πœ‹, where all four functions are exactly the same.

Example 1: Finding One-Sided Limits of a Function from Its Graph

Use the graph shown to find limο—β†’οŠ§οŽ©π‘“(π‘₯).

Answer

For the graph, the function is defined as 𝑓(π‘₯)=3 when π‘₯>1. Therefore, limο—β†’οŠ§οŽ©π‘“(π‘₯)=3, the same constant.

Be sure not to get confused with the position of the β€œ+” and especially the β€œβˆ’β€ signs in the problem. The ones appearing as superscripts on the right side indicate whether we are looking at right-sided or left-sided limits, respectively.

Example 2: Finding One-Sided Limits of a Function from Its Graph

Find limο—β†’οŠ±οŠ©οŽͺ𝑓(π‘₯).

Answer

From the graph, we see that, in spite of the value 𝑓(βˆ’3)=0, the limiting value as π‘₯β†’βˆ’3 from the left is 2: limο—β†’οŠ±οŠ©οŽͺ𝑓(π‘₯)=2.

When does a one-sided limit not exist? A simple example is provided by 𝑓(π‘₯)=1π‘₯ with its domain β„βˆ’{0}.

As we see from its graph, it makes sense to say the values 1π‘₯ become infinitely large positive as π‘₯β†’0, and we express this (and similarly, what happens as π‘₯β†’0) by writing limandlimο—β†’οŠ¦ο—β†’οŠ¦οŽ©οŽͺ1π‘₯=+∞1π‘₯=βˆ’βˆž.

In other words, there is the notion of β€œinfinite limits.”

Example 3: Finding One-Sided Limits of a Function from Its Graph

Determine limο—β†’οŠ±οŠ«οŽ©π‘“(π‘₯).

Answer

Looking at the vertical asymptote at π‘₯=βˆ’5, we see that as π‘₯ approaches this value from the right-hand side, 𝑓(π‘₯) becomes increasingly large and negative, so limο—β†’οŠ±οŠ«οŽ©π‘“(π‘₯)=βˆ’βˆž.

It is worth noting the following example, where limο—β†’οŠ±οŠ§οŽ©π‘“(π‘₯) really does not exist, while limο—β†’οŠ±οŠ§οŽͺ𝑓(π‘₯) does.

The function β€œoscillates” more and more rapidly between 1 and βˆ’1 as π‘₯β†’βˆ’1 from the right-side, so that it never settles on a limit. As π‘₯β†’βˆ’1 from the left-side, the graph indicates that the values are approaching 0.25. Finally, the actual value of this function at π‘₯=βˆ’1 is 1.1. So limdoesnotexistlimο—β†’οŠ±οŠ§ο—β†’οŠ±οŠ§οŽ©οŽͺ𝑓(π‘₯),𝑓(π‘₯)=0.25,𝑓(βˆ’1)=1.1.

In fact, the description of this function is 𝑓(π‘₯)=⎧βŽͺ⎨βŽͺ⎩π‘₯4π‘₯<βˆ’1,1.1π‘₯=βˆ’1,ο€Ό1π‘₯+1π‘₯>βˆ’1.ififsinif

In order to understand the β€œoscillation,” consider at what values π‘₯ we have sinο€Ό1π‘₯+1=0. Since sinifandonlyif(π‘₯)=0π‘₯=…,βˆ’2πœ‹,βˆ’πœ‹,0,πœ‹,2πœ‹,…, it follows that sinifandonlyifο€Ό1π‘₯=0π‘₯=…,βˆ’12πœ‹,βˆ’1πœ‹,1πœ‹,12πœ‹,13πœ‹,…, and therefore, looking near βˆ’1, but only to the right, we have sinifο€Ό1π‘₯+1=0π‘₯=βˆ’1+1πœ‹,βˆ’1+12πœ‹,βˆ’1+13πœ‹,βˆ’1+14πœ‹,βˆ’1+15πœ‹,…, which gives a sequence of points on the number line that approach βˆ’1 while the difference between successive terms gets smaller and smaller. These are the numbers where the graph above crosses the π‘₯-axis.

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