Properties of Limits

With both a conceptual and rigorous understanding of limits, it’s time to practice with them. Limits have certain properties that make computing much easier. These can be intuitively memorized, but proofs with the epsilon-delta definitions will be provided and they’re worth reading through. To be rigorous, each identity only holds true if the all limits mentioned in it exist (you’d have a hard time plugging in nonexistent values).

Note that section titles in this chapter are the names I’ve given to each property, not recognized terms. Also, most pedagogy I’ve read presents these as an enumeration of facts to memorize, but even without delta-epsilon, these can be understood very easily with a simple visualization. As an aid, figures are included.

The constant property

The first property is very simple. When dealing with a horizontal line, the limit as $x$ approaches any value is the single output value of the function. Fig. 1 illustrates a function in this form, $y = 2\text{.}$

We can formalize this observation with the notation:

Given that $c$ is a constant, the limit as $x$ approaches any real $a$ is $c\text{.}$ With a positive $\varepsilon\text{,}$ first notate $f(x) = c\text{.}$ We need to find a positive $\delta$ such that

Choosing any $\delta$ at all, $\delta = 1$ for example, $f(x)$ does not change from $c$ for all $x\text{.}$ Therefore, $|f(x) - c| = 0\text{,}$ and since $\varepsilon > 0\text{,}$ $|f(x) - c| < \varepsilon\text{.}$ $\square$

The addition property

When one function’s outputs are defined as the sum of the outputs of two other functions, the limit of this function at any input value is the sum of the limits of the two functions. Fig. 2 illustrates this with an orange and green function, as well as their sum in dashed green.

Since the limits exist at $x = 4\text{,}$ every output near that x-value behaves as if the hole didn’t exist. You can think of this as simply adding the existing surrounding output values, which would give the dashed green function a hole at the sum of the two others. This property can be formally notated as

To prove this, we’ll use the delta-epsilon definition of limits again. First notate each limit separately:

Now for any $\varepsilon > 0\text{,}$ we must prove that there exists a $\delta > 0$ such that

We can now split the desired inequality into two terms and prove each one below $\varepsilon / 2\text{.}$ First, $\lim\limits_{x \to a} f(x) = L$ means there exists a $\delta_1$ such that

The other limit, $\lim\limits_{x \to a} g(x) = M\text{,}$ means there exists a $\delta_2$ such that

If we now let $\delta = \min(\delta_1, \delta_2)\text{,}$ we will have both $|f(x) - L|$ and $|g(x) - M|$ less than $\varepsilon / 2\text{.}$ Rearrange our desired inequality to get terms we want:

By the triangle inequality:

Since we have demonstrated that for every $\varepsilon > 0\text{,}$ there exists a $\delta = \min(\delta_1, \delta_2) > 0$ such that

proving the property. $\square$

The product property

The next property is that the limit of the product of two functions is the product of the limits of each function. Take a look at Fig. 3, which shows two functions and their product.

The blue function is linear, the orange function is exponential, and the dashed green function is the product of the first two. As indicated by the open circles, the blue and orange functions are undefined for $x = 2\text{.}$ Therefore, the function defined by the product of their outputs also is undefined at $x = 2\text{.}$

To understand this property intuitively, you can find the limit as $x$ approaches $2$ of the green function by multiplying the approached output values of the orange and blue functions, since each graph functions as if no hole existed in the surrounding domain, excluding $2$ itself.

Expressing this property formally in notation:

To prove this, first let

We want to prove for any $\varepsilon > 0$ that there exists a $\delta$ such that

Right now, we’re working with

and we want to introduce expressions for each of our separate limits. To do this, add a net-zero expression.

Now we factor:

By the triangle inequality we know

The expression on the right is equivalent to

Our goal is now to prove each term to be less than $\varepsilon / 2\text{,}$ because this necessitates that their sum is under $\varepsilon\text{.}$ The factor of $|g(x)|$ in the first term is a complication, because we can’t guarantee predict both how $|f(x) - L|$ and $|g(x)|$ behave at the same time in order to control their product. By the delta-epsilon definition, setting $\varepsilon = 1\text{,}$ there exists a $\delta_1 > 0$ such that

By triangle inequality:

Our goal is

Substitute $|g(x)| < |M| + 1$ to get

To prove this true, start with the delta-epsilon definition of a limit. $\lim\limits_{x \to a} f(x) = L$ means there exists a $\delta_2 > 0$ such that

Since we now have

we multiply both sides by $|g(x)| < |M| + 1$ (since both terms on the left are nonnegative) and the following inequality is proven true:

Our next goal is proving

This is simpler since $|L|$ is a constant. However, we now have to consider the case $L = 0\text{,}$ which prohibits division. Case A will assume $L \neq 0$ and Case B will assume $L = 0\text{.}$

In Case A, we want

Now, by definition of $\lim\limits_{x \to a} g(x) = M\text{,}$ there exists a $\delta_3 > 0$ such that

Multiplying both sides by $|L|$ yields our desired result of

In Case B, we have $L = |L| = 0\text{.}$ Given that $\varepsilon > 0\text{,}$ the following is trivial:

Now that we’ve covered $\delta_1, \delta_2, \delta_3$ such that if $x$ is less than all of them, we can guarantee

Let $\delta = \min(\delta_1, \delta_2, \delta_3)$ in order to satisfy all tolerances. Then for all $0 < |x - a| < \delta$ it is true that

which concludes the proof. $\square$

The exponent property

Note that from the product property, we can get another property for free:

Intuitively, the limit of any function to a power $n$ is equal to the limit of the function raised to $n\text{.}$ This is because raising the function to the power of $n + 1$ is the same as multiplying the function by the function raised to the power of $n\text{.}$ A rigorous proof is easily done through induction.

The base case is $n = 1\text{.}$ Raising anything to the power of $1$ leaves it unchanged:

Assume now that the property holds true for $n\text{.}$ This means that

Apply the product property with the factor $\lim\limits_{x \to a} f(x)$ to both sides:

Which means the formula holds for $n + 1$ given that it does for $n\text{.}$ Since we’ve proven the base case $n = 1\text{,}$ the property holds for all positive integers $n\text{.}$ $\square$

The scalar multiple property

With the product and constant properties proven, the next property can be done without a separate delta-epsilon proof. This property states that the limit of a function multiplied by a constant is the constant multiplied by the limit of the function. In notation, let $c$ be a constant and $\lim\limits_{x \to a} f(x) = L\text{.}$ The scalar multiple property states that

By the product property, we know

The constant property states that $\lim\limits_{x \to a} c = c\text{,}$ so substitution yields

and the property is proved. $\square$

The reciprocal property

This property has a self-explanatory name. The limit of a function’s reciprocal is the reciprocal of the limit. Let $\lim\limits_{x \to a} f(x) = L\text{.}$ Obviously, since $L$ appears in the denominator, this property assumes $L \neq 0\text{.}$

By delta-epsilon, we need to prove that for any $\varepsilon > 0\text{,}$ there exists a $\delta > 0$ such that

We can find a common denominator:

Now we would like to bound the $|f(x)|$ in the denominator. Since we know $\lim\limits_{x \to a} f(x) = L\text{,}$ we can use the delta-epsilon with an epsilon defined as half of $|L|\text{:}$

By the triangle inequality:

Since we know $|L - f(x)| = |f(x) - L| < \frac{|L|}{2}$

Substituting this inequality into our expression yields

Recall that we wanted this value to be less than $\varepsilon\text{:}$

Because $\lim\limits_{x \to a} f(x) = L\text{,}$ we can simply set the vertical tolerance to $\frac{\varepsilon |L|^2}{2}\text{,}$ meaning that there must exist a $\delta_2$ such that

If we now set $\delta = \min(\delta_1, \delta_2)\text{,}$ both conditions hold:

Now substitute into the original expression:

which completes the delta-epsilon proof. $\square$

The quotient property

This is the last limit property covered in this chapter. The product and quotient properties have similar structures, with the latter stating that the limit of the quotient of two functions is equal to the quotient of the limits of each function. Of course, this assumes that the denominator is nonzero.

Intuitively, you can understand this in the same way. Since every output value surrounding the hole functions normally, a function defined as the quotient of two others will have its hole in an appropriate place. In notation, for $\lim\limits_{x \to a} f(x) = L$ and $\lim\limits_{x \to a} g(x) = M\text{:}$

This comes trivially since we’ve already established the product and reciprocal rules. We know, by the basic rules of fractions:

By the product property:

Applying the reciprocal property:

which proves the quotient property. $\square$

In this chapter, we covered the bulk of what one needs to know when manipulating limits. Next, we’ll be contextualizing the limit with the concept of continuity, which is perhaps the most important thing the limit section has to offer. That’s all for the properties of limits. Zai jian.