In this explainer, we will learn how to use the properties of limits such as the limits of sums, differences, products, and quotients of functions and the limit of composite functions.

The concepts of limits and convergence are two of the staple ideas that form the basis of analysis, which is one of the most central ideas in mathematics and appears without fail in every university-level course in mathematics. Often, this begins with the so-called epsilon definition of convergence, which turns vague and intuitive notions into robust mathematical ideas that can be rigorously defined.

Much as an understanding of the epsilon definition is fundamental for developing related theorems and other results, in practical terms, an understanding of limits generally does not require the same level of detail. Pleasingly, operating using limits can often be summarized into a neat series of helpful results, many of which bear a strong resemblance to the normal rules of algebra that we are already practiced in. In particular, we will work with the algebra of limits theorems, which drastically undercut the amount of conceptual knowledge that is needed to understand practical examples. In the following definition, we will give a selection of results that will be of immediate use in the examples in this explainer.

It is helpful at this stage to preemptively introduce a paradigm for recalling the main results from the algebra of limits theorems. In this case, a written word version is actually fairly helpful. As we will shortly see, it is surprisingly meaningful to remember phrases such as βthe limit of the sum is the sum of the limits,β or more sophisticated expressions such as βthe composition of the limits of the limit of the compositions.β These do not necessarily make any sense but can be a useful prompt for recalling which options are available when working with the expressions involving multiple limits.

### Definition: Properties of Limits

Suppose that and that these functions are continuous around the point . Then, the following results hold true for addition and subtraction:

An analogous result holds for multiplication and is stated as follows:

Furthermore, if we suppose that , then we have the result

Additionally, if is some real constant, then we have the two results and

The final result that we will use relates to the composition of a function and is expressed as follows:

For this final result to hold, the function must be continuous around .

Please note that it is more common to refer to these collection of results as the βalgebra of limitsβ theorems, which is how we shall refer to these through the rest of the document.

This is a comprehensive set of results to introduce an all-in-one definition, but the interpretation and usage of these are simpler than might initially seem obvious. With practice, one tends to use the above results without much further thought or consideration. These results are nowhere near as intimidating as they might first seem, and in reality we can more or less just remember the normal rules of algebra and then apply these directly to limits. Proving each of the results is obviously more difficult and it relies on a solid understanding of the epsilon definition, which is outside the scope of this explainer.

We will demonstrate the above results through the examples contained in this explainer. Before we do this, we should note how several of the above results will extend in a more general sense to more than two functions. For example, suppose that three functions , , and are all continuous when and that

Then it is still the case that the limit of the sum is the sum of the limits, and that the limit of the difference is the difference of the limits. Translated into actual mathematics, this means that

Likewise, we would also find that the limit of the product is the product of the limits. In other words, we have the following result:

A more general statement of this result would involve any number of particular limits and would hold true in the obviously extensible way. Now we will apply all of thse results to the following examples.

### Example 1: Evaluating Limits Involving Sums and Differences of Functions

Given that , , and , find .

### Answer

We will invoke the algebra of limits theorems in order to solve this problem. We recall the result that the limit of the sum is the sum of the limits. Interpreted for this particular example, the result is that

We already were given these individual limits in the statement of the question, which means that we can now give the result

The algebra of limits results are so powerful that they can easily be extended to examples which appear to be highly convoluted. Just as we might work seamlessly with the majority of conventional algebraic expressions, essentially, there are very few changes to these processes when limits are introduced. Providing that we are never attempting to divide by zero (which is not thought of highly in the mathematical community), there are few consequences for treating problems exactly in this way. It can be instructional to include more steps than are strictly necessary when evaluating expressions involving multiple limits, and in the following example we will slightly overdo the number of steps that are required. Until these techniques are mastered, it is safer to write out more steps than are required, as we will shortly demonstrate.

### Example 2: Evaluating Limits Involving Products and Differences of Functions

Assume that , , and . Find .

### Answer

The algebra of limits says that the limit of the difference is the difference of the limits. Therefore we can write the original expression as

The question states that , which implies that

Now all that remains is to evaluate the remaining expression involving the limits. For this, we can recall that the limit of the products is the product of the limits, which allows us to write

We already know that and that , which means that we can evaluate

So far in this explainer, we have seen several examples of applications involving the algebra of limits. Ultimately, these problems can be thought of mostly as evaluating an algebraic expression which just so happens to involve limits. This way of thinking allows the standard framework of algebra to be employed without much additional thought or consideration. With this established, it is possible to solve equations in order to find limits that are not given. As a very simple example, suppose we were told that and that

If we were asked to calculate the limit of when , then we could apply the particular result from the algebra of limits which states that the limit of the sum is the sum of the limits. This allows us to rewrite the equation above as

Given that we were told the limit of when is equal to 3, we have that which immediately rearranges to give

This principle of solving equations featuring limit terms is extended in the natural way using the other results from the algebra of limits. We will demonstrate this in the following example.

### Example 3: Evaluating Limits Involving Quotients and Constant Multiple of Functions

Given that , determine .

### Answer

We will begin by creating an auxiliary function , which allows us to write the original expression in the following form:

Additionally, we evaluate the limit of as , which clearly gives

We can now apply the algebra of limits results directly. In particular, we recall that the limit of the quotient is the quotient of the limit. This means that, providing that the limit of the denominator term is not equal to zero, we can rewrite equation (1) as

Rearranging this expression gives

Given equation (2), we can now give the final result:

The trick of introducing an auxiliary function is one that is not required when answering such questions, but it has the helpful effect of phrasing the question in terms of the results from the algebra of limits. Given the range of results from this collection of theorems, it is often possible to answer a question using more than one method. For example, in the previous question, we could have recalled the general result that for any constant . This would have allowed us to make the equivalence from which we could have solved the problem using similar steps to the previous example. Further recognition that could also have allowed this term to effectively be treated as a constant.

### Example 4: Evaluating Limits Involving Differences and Roots of Functions

Assume that and . Find .

### Answer

It will initially be helpful to rephrase the limit above in terms of slightly different notation as

Given the algebra of limits theorems, we know that

We can now recall that the limit of the difference is the difference of the limits, which means that we can write

We were told in the statement of the question that and , which means that the above expression can be simplified as

In the next example, we will use the properties of limits to evaluate limits involving composite functions.

### Example 5: Evaluating Limits Involving Composite Functions

Assume that . If , find .

### Answer

The relevant result from the algebra of limits theorems is as follows. Assume that there are two functions and such that and that the function is continuous when . Then we have

Applying this to the question above, we would find

We have seen during this explainer that the algebra of limits results can be used in a fairly cavalier way once their usage is properly understood. Providing we are comfortable with the normal rules of algebra, introducing limits into equations does not often provide much in the way of complications. That said, it is always advisable to solve such problems while annotating working with the results from the algebra of limits that have been used to progress the solution. In our final example, we will combine several of the techniques that we have previously covered in this explainer, while extending these to cover functions which are defined graphically. We have given little attention to the conditions that we should impose on functions in order for the algebra of limits results to apply. In essence, we require all functions involved to be continuous in the region of the limit that we are interested in. Expressed graphically, we require that the plot of a function is smooth around this limit, although not necessarily in other regions.

### Example 6: Evaluating Limits Involving Product and Powers of Functions Which Are Represented Graphically

Consider the graph of .

Find .

### Answer

The graph above is not smooth in all regions, which means that the function is not continuous for all real numbers. Indeed, there is even a gap in the plot of the graph when takes values between and . Nevertheless, the function is smooth in the region of , which means that we can assume that the function is continuous at this point. This means that we can evaluate by reading off the value of the function at this point.

We can see from the plot above that

Now we will begin to evaluate the limit as given in the question. We know that the exponent of 2 can be moved in order to simplify the calculations by using one of the results from the algebra of limits, in particular, as follows:

Now, we can recall another result from the algebra of limits which says that the limit of the product is the product of the limits. In other words, we can simplify the previous expression as

We can easily see that and we have the result from equation (3), both of which allow us to simplify the previous expression as follows:

The algebra of limits results are some of the most simple and powerful tools we have when trying to evaluate the limits of composite or nontrivial functions. Once the epsilon definition has been understood, proving the algebra of limits results from this is a valuable experience for any student who is new to this subject. Once these results have been proven (e.g., as part of a standard university course on analysis), they can be applied in a relatively thoughtless way, providing that precautions are taken (such as ensuring that we are never dividing by a function which has a limit of zero at the point of interest). Generally, we will be working with functions that are continuous for every element of their domain and so this is not likely to be an issue. However, if we are given a piecewise function (such as the one in the previous question), then the matter is more delicate.

### Key Points

Suppose that and are two functions that are continuous around . Further assume that is a real constant and that and , where . Then, the following results apply from the algebra of limits theorems:

It can be useful to apply the following expressions to each of these results where possible: βthe limit of the sum is the sum of the limitβ and βthe limit of product is the product of the limits.β