Lesson Part 1

Recall

To this point, we have seen that the graph of \(f(x)\) can be very helpful in evaluating \(\displaystyle \lim_{x\to a} f(x)\).

But, what if we wanted to evaluate this limit algebraically without the aid of the graph?

The following properties of limits can be used to algebraically evaluate a limit.

Properties of Limits

Here is a list of properties of limits.

If \(\displaystyle \lim_{x\to a} f(x) \) and \(\displaystyle \lim_{x \to a} g(x)\) exist, then we have the following properties:

Suppose that we are evaluating the limit as \(x\) approaches \(a\) of a function where the function is a constant. For example, \(k\) might be the number \(5\). The value of this limit is also \(5\). So when we are evaluating the limit as \(x\) approaches \(a\) of a constant \(k\), the value of that limit is \(k\).

When evaluating the limit as \(x\) approaches \(a\) of function \(x\), the value of the limit is \(a\).

If we are adding two functions together, \(f(x) + g(x)\), and we are asked to evaluate the limit as \(x\) approaches \(a\), then the limit of \(f(x) + g(x)\) is the same as calculating the limit of each function separately and then adding those values together. So here we see that the limit as \(x\) approaches \(a\) of \(f(x) + g(x)\) is the same as the limit as \(x\) approaches \(a\) of \(f(x)\) plus the limit as \(x\) approaches \(a\) of \(g(x)\). This is also the case when we are subtracting \(g(x)\) from \(f(x)\).

If we are multiplying a function times a constant, in this case we're using \(c\) to represent that constant, we see that the limit as \(x\) approaches \(a\) of a constant \(c\) times \(f(x)\) is the same as multiplying the value of the limit as \(x\) approaches \(a\) of \(f(x)\) by \(c\). In other words, if we are multiplying by a constant inside the limit, we can remove that constant from inside the limit and multiply by it after we've evaluated the limit on \(f(x)\).

If we have a product of two functions, in this case, we're multiplying \(f(x)\) times \(g(x)\), and we are asked to find the limit as \(x\) approaches \(a\), then the value of this limit is equal to finding the limit of each function separately and then multiplying these two values together. So in this case, the limit as \(x\) approaches \(a\) of \(f(x)\) times \(g(x)\) is the same as the limit as \(x\) approaches \(a\) of \(f(x)\) times the limit as \(x\) approaches \(a\) of \(g(x)\).

If we have a quotient of two functions, say \(f(x)\) divided by \(g(x)\), and we're asked to evaluate the limit as \(x\) approaches \(a\), the value of this limit is equal to calculating the limit of both functions separately and then dividing their two values. So here we write this as the limit as \(x\) approaches \(a\) of \(f(x)\) divided by \(g(x)\) is equal to the limit as \(x\) approaches \(a\) of \(f(x)\) divided by the limit as \(x\) approaches \(a\) of \(g(x)\). Since the limit as \(x\) approaches \(a\) of \(g(x)\) is in the denominator, the value of this limit cannot be zero.

If we are asked to evaluate the limit as \(x\) approaches \(a\) of a function and on that function is an exponent, the value of this limit is the same as evaluating the limit as \(x\) approaches \(a\) of \(f(x)\) and then applying the exponent after we have evaluated this limit. It should be noted here that this works in most cases but not always.
Note: If \(r\) is negative, then we are assuming \(\displaystyle \lim_{x\to a} f(x) \neq 0\). If \(r=\frac{p}{q}\) in lowest terms with \(q\) even, then we are assuming that \(\displaystyle \lim_{x\to a} f(x) \geq 0\). For example, if \(r\) is equal to \(\dfrac{1}{2}\), and the limit of \(f(x)\) is a negative number, then we can't calculate the square root of this negative limit. So we need to make sure that we take extra care when applying limit property seven.

1. \(\displaystyle \lim_{x\to a} k = k\), where \(k\) is a constant.
2. \(\displaystyle \lim_{x\to a} x = a\).
3. \(\displaystyle \lim_{x\to a} \left [ f(x)\pm g(x) \right ] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x) \).
4. \(\displaystyle \lim_{x\to a} cf(x) = c \left[ \lim_{x\to a} f(x) \right]\).
5. \(\displaystyle \lim_{x\to a} \left [ f(x)g(x)\right ] = \left [ \lim_{x\to a} f(x) \right ] \left [ \lim_{x\to a} g(x) \right ]\).
6. \(\displaystyle \lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)}\), provided \(\displaystyle\lim_{x\to a} g(x) \neq 0\).
7. \(\displaystyle \lim_{x\to a} \left [ f(x) \right ]^{~r} = \left [ \lim_{x\to a} f(x) \right ]^r\), for most real numbers, \(r\).

Lesson Part 2

Examples

Example 1

Let's now try the following example. Consider the limit \(\displaystyle L=\lim_{x\to 2} \left ( x^2+5x-7\right )\). Look at the list of the limit properties and spot the limit property that we can apply to evaluate this limit.

If \(\displaystyle \lim_{x\to a} f(x) \) and \(\displaystyle \lim_{x \to a} g(x) \) exist, then we have the following properties

1. \(\displaystyle \lim_{x\to a} k = k\), where \(k\) is a constant.
2. \(\displaystyle \lim_{x\to a} x = a\).
3. \(\displaystyle \lim_{x\to a} \left [ f(x)\pm g(x) \right ] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x) \).
4. \(\displaystyle \lim_{x\to a} cf(x) = c \left[ \lim_{x\to a} f(x) \right]\).
5. \(\displaystyle \lim_{x\to a} \left [ f(x)g(x)\right ] = \left [ \lim_{x\to a} f(x) \right ] \left [ \lim_{x\to a} g(x) \right ]\).
6. \(\displaystyle \lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)}\), provided \(\displaystyle\lim_{x\to a} g(x) \neq 0\).
7. \(\displaystyle \lim_{x\to a} \left [ f(x) \right ]^{~r} = \left [ \lim_{x\to a} f(x) \right ]^r\), for most real numbers, \(r\).
   Note: If \(r\) is negative, then we are assuming \(\displaystyle \lim_{x\to a} f(x) \neq 0\). If \(r=\frac{p}{q}\) in lowest terms with \(q\) even, then we are assuming that \(\displaystyle \lim_{x\to a} f(x) \geq 0\).

Solution

The first limit property that we will apply is limit property three, which tells us when we are evaluating a limit that has addition and subtraction contained within the function, we can evaluate the limit of each term in the addition and subtraction sentence separately, and then add and subtract the re-evaluated amounts.

\[L = \displaystyle \lim_{x\to 2} x^2 + \lim_{x\to 2} 5x - \lim_{x\to 2} 7 \]

Now that we have applied limit property three, consider each of these limits separately. Let's look at the first limit. Which limit property should we apply to the limit as \(x\) approaches \(2\) of \(x\) all raised to the exponent \(2\)? Limit properly seven talks about a function raised to an exponent. So we can apply limit property seven here to change the limit as \(x\) approaches \(2\) of \(x^2\), to become the limit as \(x\) approaches \(2\) of \(x\) all raised to the exponent \(2\). Moving to the second limit, we see that we have a constant times a function of \(x\). Limit property four tells us that we can remove the constant multiple of \(5\) from the limit and evaluate the limit on \(x\). So the limit as \(x\) approaches \(2\) of \(5x\) becomes \(5\) times the limit as \(x\) approaches \(2\) of \(x\). The last limit is a limit of a constant. We will use limit property one in the next line to evaluate this limit.

\[\displaystyle \lim_{x\to 2} x^2 + \lim_{x\to 2} 5x - \lim_{x\to 2} 7 =\displaystyle \left (\lim_{x\to 2} x \right )^2 +5 \left ( \lim_{x\to 2} x \right ) - \lim_{x\to 2} 7 \]

Now that we have applied limit properties three, four, and seven, we can use limit properties one and two to finally evaluate this limit. So for the limit as \(x\) approaches \(2\) of \(x\) we will apply limit property two, which tells us that the value of this limit is the value that is approached. So we will substitute the value of \(2\) for the limit. When we look at the second limit, we notice that the limit inside the brackets is equal to the limit that we just evaluated. Again, we will substitute \(2\) in for the limit. The last limit is just the limit of a constant. And limit property one tells us that the limit of a constant evaluates to the constant itself.

\[\displaystyle \left (\lim_{x\to 2} x \right )^2 +5 \left ( \lim_{x\to 2} x \right ) - \lim_{x\to 2} 7 = (2)^2 +5(2)-7 \]\[(2)^2 +5(2)-7 = 7\]

Putting it all together, we have

\[\begin{align*}L &= \displaystyle \lim_{x\to 2} x^2 + \lim_{x\to 2} 5x - \lim_{x\to 2} 7 &\text{ by property 3}\\ L &=\displaystyle \left (\lim_{x\to 2} x \right )^2 +5 \left ( \lim_{x\to 2} x \right ) - \lim_{x\to 2} 7 &\text{ by properties 7 and 4}\\ L &= (2)^2 +5(2)-7 &\text{by properties 1 and 2}\\ L &= 7\end{align*}\]

Example 2

Now try the next example on your own, and then we will look at which limit properties we will apply and the order at which they will be applied.

Evaluate \(\displaystyle L = \lim_{x\to -1} \dfrac{2x+3}{x-5}\).

Solution

When we look at the function \(\dfrac{2x + 3}{x - 5}\), we see a quotient. So the very first limit property to apply is limit property six.

\[L = \dfrac{\displaystyle \lim_{x\to -1} (2x+3)}{\displaystyle \lim_{x\to -1} (x-5)} \]

We now have a limit in the numerator and a separate limit in the denominator. So we look at these on their own. Let's consider the numerator. The numerator is the addition of two functions. We know, then, to apply limit properly three. In the denominator, we have the difference of two functions. So again, we will apply limit property three.

\[\dfrac{\displaystyle \lim_{x\to -1} (2x+3)}{\displaystyle \lim_{x\to -1} (x-5)} = \dfrac{\displaystyle \lim_{x\to -1} 2x + \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} \]

Now we have four separate limits. Let's begin with the first term in the numerator. In this case, we have a constant multiple of \(2\) applied to the function \(x\). Limit property four tells us that we can remove the \(2\) from inside the limit. In the second term of the numerator, we see that it's a limit of a constant. We will deal with that on our next line. In the denominator, we have the limit of the function \(x\), and we also have the limit of a constant. Let's wait till the next line to evaluate these two limits.

\[\dfrac{\displaystyle \lim_{x\to -1} 2x + \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} = \dfrac{\displaystyle 2\left (\lim_{x\to -1} x \right )+ \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} \]

Since we are left with only the limit as \(x\) approaches \(-1\) of \(x\), or the limit as \(x\) approaches \(-1\) of a constant, we're going to apply limit properties one and two to evaluate this limit.

\[\dfrac{\displaystyle 2\left (\lim_{x\to -1} x \right )+ \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} = \dfrac{2(-1)+3}{(-1)-5} \]\[\dfrac{2(-1)+3}{(-1)-5} = -\dfrac{1}{6}\]

You should arrive at the answer \(-\dfrac{1}{6}\). Putting it all together, we have

\[\begin{align*}L &= \dfrac{\displaystyle \lim_{x\to -1} (2x+3)}{\displaystyle \lim_{x\to -1} (x-5)} &\text{by property 6}\\ L &= \dfrac{\displaystyle \lim_{x\to -1} 2x + \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} &\text{by property 3}\\ L &= \dfrac{\displaystyle 2\left (\lim_{x\to -1} x \right )+ \lim_{x\to -1} 3}{\displaystyle \lim_{x\to -1} x -\lim_{x\to -1} 5} &\text{by property 4}\\ L &= \dfrac{2(-1)+3}{(-1)-5} &\text{by properties 1 and 2}\\ L &= -\dfrac{1}{6}\end{align*}\]

Challenge Question

Now that we've done some practice with the first and second example, we now will know how to evaluate the limit of any polynomial or any rational function. The method of applying the limit properties is the same for any polynomial or rational function. Let's now try the following challenge question together.

Evaluate \(\displaystyle L=\lim_{x\to 16} \dfrac{(2x-5)^{\frac{1}{3}}}{x-7} \).

Solution

When we look at the function inside the limit, we see a quotient, but we also see that the numerator is raised to the rational exponent \(\dfrac{1}{3}\). What limit properties will we apply to evaluate this limit? The first limit property that we should use is limit property six. Since we have a quotient of two functions, applying limit property six gives us a limit in the numerator that's separate from the limit of the denominator.

\[L = \dfrac{\displaystyle\lim_{x\to 16} \left [ (2x-5)^{\frac{1}{3}}\right ]}{\displaystyle\lim_{x\to 16} \left [ x-7 \right ]} \]

Now let's consider the numerator. The function in the numerator has the rational exponent \(\dfrac{1}{3}\). So we can apply limit properly seven to remove this rational exponent from inside the limit.

\[L = \dfrac{\displaystyle \left [ \lim_{x\to 16} (2x-5)\right ]^{\frac{1}{3}}}{\displaystyle \lim_{x\to 16} \left [ x-7\right ]} \]

In the denominator, we simply have a subtraction. In this case, we can apply limit property three to evaluate the limit of each term separately. We can also apply limit property three to the numerator since once the rational exponent has been removed from the function, we see that we still have a subtraction of two terms.

\[L = \dfrac{\displaystyle \left [ \lim_{x\to 16} 2x-\lim_{x\to 16} 5 \right ]^{\frac{1}{3}}}{\left[ \displaystyle \lim_{x\to 16} x - \displaystyle \lim_{x\to 16} 7 \right]}\]

Now, the first limit in the numerator has a constant multiple of \(2\) applied to \(x\). Limit property four tells us that we can remove this constant multiple from inside the limit.

\[L = \dfrac{\displaystyle \left [ 2\left (\lim_{x\to 16} x \right )-\lim_{x\to 16} 5 \right ]^{\frac{1}{3}}}{\left [ \displaystyle \lim_{x\to 16} x -\lim_{x\to 16} 7 \right ]} \]

Once we have applied limit properties six, seven, three, and four, we are now ready to evaluate this limit by applying limit properties one and two, which leaves us with

\[L = \dfrac{\left [ 2(16)-5 \right ]^{\frac{1}{3}}}{[(16)-7]} \]\[L = \dfrac{(27)^{\frac{1}{3}}}{9}\]\[L = \dfrac{3}{9}\]\[L = \dfrac{1}{3}\]

Putting it all together, we have

\[\begin{align*}L &= \dfrac{\displaystyle\lim_{x\to 16} \left [ (2x-5)^{\frac{1}{3}}\right ]}{\displaystyle\lim_{x\to 16} \left [ x-7 \right ]} &\text{by property 6}\\ L &= \dfrac{\displaystyle \left [ \lim_{x\to 16} (2x-5)\right ]^{\frac{1}{3}}}{\displaystyle \lim_{x\to 16} \left [ x-7\right ]} &\text{by property 7}\\ L &= \dfrac{\displaystyle \left [ \lim_{x\to 16} 2x-\lim_{x\to 16} 5 \right ]^{\frac{1}{3}}}{\left[ \displaystyle \lim_{x\to 16} x - \displaystyle \lim_{x\to 16} 7 \right]} &\text{by property 3}\\ L &= \dfrac{\displaystyle \left [ 2\left (\lim_{x\to 16} x \right )-\lim_{x\to 16} 5 \right ]^{\frac{1}{3}}}{\left [ \displaystyle \lim_{x\to 16} x -\lim_{x\to 16} 7 \right ]} &\text{by property 4}\\ L &= \dfrac{\left [ 2(16)-5 \right ]^{\frac{1}{3}}}{[(16)-7]} &\text{by properties 1 and 2}\\ L &= \dfrac{(27)^{\frac{1}{3}}}{9}\\ L &= \dfrac{3}{9}\\ L &= \dfrac{1}{3}\end{align*}\]

Quiz

See the quiz in the side navigation.