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# 8.2: Computing Limits

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Demonstrate an understanding of the limit of a function from an algebraic perspective.
• Learn how to compute the limit for different kinds of functions.

## Introduction

In this lesson, we will present the limit of a function from an algebraic perspective. We will base the algebraic technique on the results obtained from the previous section. We first present the basic theorems that provide the tools necessary to calculate limits. These theorems are outlined in the two boxes below.

Box #1: Important Theorems of Limits

Let a be a real number and suppose that \begin{align*}\lim_{x\rightarrow a}f(x) = L_1 \end{align*} and \begin{align*}\lim_{x\rightarrow a}g(x) = L_2 \end{align*}.

Then:

1. \begin{align*} \lim_{x\rightarrow a} [f(x) + g(x)] = \lim_ {x\rightarrow a}f(x) + \lim_{x\rightarrow a}g(x) = L_1 + L_2\end{align*}, meaning the limit of the sum is the sum of the limits.

2. \begin{align*} \lim_{x\rightarrow a} [f(x) - g(x)] = \lim_ {x\rightarrow a}f(x) - \lim_{x\rightarrow a}g(x) = L_1 - L_2\end{align*}, meaning the limit of the difference is the difference of the limits

3. \begin{align*} \lim_{x\rightarrow a} [f(x)g(x)] = ( \lim_ {x\rightarrow a}f(x)) ( \lim_{x\rightarrow a}g(x)) = L_1L_2\end{align*}, meaning the limit of the product is the product of the limits.

4. \begin{align*} \lim_{x\rightarrow a} \frac{f(x)}{g(x)} =\frac{\lim_{x\rightarrow a} f(x)} {\lim_{x\rightarrow a} g(x)} = \frac {L_1} {L_2} L_2 \neq 0\end{align*}, meaning the limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero.)

5. \begin{align*} \lim_{x\rightarrow a} \sqrt[n]{f(x)}= \sqrt[n] { \lim_{x\rightarrow a} f(x)} = \sqrt[n] L_1 L_1 > 0\end{align*} if n is even, meaning the limit of the nth root is the nth of the limit.

Other useful results follow from the above theorems:

Box #2

1. If a and k are real numbers, then \begin{align*}\lim_{x\rightarrow a}k = k\end{align*}. That is, if f(x) = k, a constant function, then the values of f(x) do not change as x is varied, thus the limit of f(x) is k.

2. If a is a real number then \begin{align*}\lim_{x\rightarrow a} x= a \end{align*}. That is, since f(x) = x is an identity function (its input equals its output), then as xa, f(x) = xa.

3. \begin{align*}\lim_{x \rightarrow a} (k \cdot f(x)) = ( \lim_{x \rightarrow a} k) \cdot (\lim_{x \rightarrow a} f(x)) = k \cdot ( \lim_{x \rightarrow a} f(x))\end{align*}

4. \begin{align*}\lim_{x \rightarrow a} x^n = ( \lim_{x \rightarrow a} x)^n = a^n\end{align*}

## Limits of Polynomial Functions

Example 1: Find \begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4)\end{align*} and justify each step.

Solution: Using Equation (1) of Box #1 (the limit of the sum is the sum of the limits) we get

\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = \lim_{x \rightarrow 1}x^2 + \lim_{x \rightarrow 1}(-3x) + \lim_{x \rightarrow 1} 4\end{align*}
From Equation (4) of Box #2, the first term becomes
\begin{align*}\lim_{x \rightarrow 1}x^2 = (1)^2 = 1\end{align*}
From Equations (2) and (3) of Box #2, the second term becomes
\begin{align*}\lim_{x \rightarrow 1}(-3x) = -3 \lim_{x \rightarrow 1}x = (-3)(1) = -3\end{align*}
Finally, from Equation (1) of Box #2, the third term becomes
\begin{align*}\lim_{x \rightarrow 1}4 = 4\end{align*}

Thus the limit of the above polynomial is

\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = 1 + (-3) + 4 = 2\end{align*}

However, for conciseness, we can compute the limit by simply substituting x = 1 directly into the polynomial,

\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = (1)^2 - 3(1) + 4 = 2\end{align*}

Theorem: The limit of a polynomial

For any polynomial f(x) = cnxn + . . . + c1x + c0 and any real number a,\begin{align*}\lim_{x \rightarrow a}f(x) = c_n(a)^n + . . . + c_1(a) + c_0\end{align*}

\begin{align*}\lim_{x \rightarrow a}f(x) = f(a)\end{align*}

In other words, the limit of the polynomial is simply equal to f(a).

Example 2: Find \begin{align*}\lim_{x\rightarrow 3} (4x^3 - 4x - 5)\end{align*}

Solution: According to the theorem above, \begin{align*}\lim_{x\rightarrow 3}(4x^3 - 4x - 5) = 4(3)^3 - 4(3) - 5 =91\end{align*}

Example 3: Find \begin{align*}\lim_{x\rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} \end{align*}

Solution: Using Equation (4) of Box #1 (the limit of the quotient is the quotient of the limit),

\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} = \frac {\lim_{x \rightarrow 5} (x^2 - 4)} {\lim_{x \rightarrow 5}(3x^2 - 2)}\end{align*}
Making use of the limit of the polynomials theorem, we obtain,
\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2}\end{align*} \begin{align*}= \frac{(5)^2 - 4} {3(5)^2 - 2}\end{align*}
\begin{align*}\mathit = \frac{21} {73}\end{align*}

## Limits of Rational Functions

In Example 3 we found the limit of a rational function. Sometimes finding the limit of a rational function at a point a is difficult because evaluating the function at the point a leads to a denominator equal to zero. The box below describes finding the limit of a rational function.

Theorem: The limit of a Rational Function

For the rational function \begin{align*}f(x) = \frac{p(x)} {q(x)}\end{align*} and any real number a,

\begin{align*}\lim_{x \rightarrow a} f(x) = \frac{p(a)} {q(a)} \text{ if } q(a) \neq 0\end{align*}.

However, if \begin{align*} \,\! q(a) = 0 \end{align*} then the function may or may not exist. See Examples 5, 6, and 7 below.

Example 4: Find \begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2}\end{align*}.

Solution: Using the theorem above, we simply substitute x = 3: \begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2} = \frac{2 - 3} {3 - 2} = -1\end{align*}

Example 5: Find \begin{align*}\lim_{x \rightarrow 3} \frac{x + 1} {x - 3}\end{align*}.

Solution: Notice that the domain of the function is continuous (defined) at all real numbers except at x = 3. If we check the one-sided limits we see that \begin{align*}\lim_{x \rightarrow 3^{+}} \frac{x + 1} {x - 3}=\infty\end{align*} and \begin{align*}\lim_{x \rightarrow 3^{-}} \frac{x + 1} {x - 3}=-\infty\end{align*}. Because the one-sided limits are not equal, the limit does not exist.

Example 6: Find \begin{align*}\lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2}\end{align*}.

Solution: Notice that the function here is discontinuous at x = 2, that is, the denominator is zero at x = 2. However, it is possible to remove this discontinuity by canceling the factor x-2 from both the numerator and the denominator and then taking the limit:

\begin{align*}\lim_{x \rightarrow 2} \frac{x^2 -4} {x - 2} = \lim_{x \rightarrow 2} \frac{(x - 2)(x + 2)} {x - 2} = \lim_{x \rightarrow 2} (x + 2) = 4\end{align*}

This is a common technique used to find the limits of rational functions that are discontinuous at some points. When finding the limit of a rational function, always check to see if the function can be simplified. We will use this technique again in the next example.

Example 7: Find \begin{align*}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align*}.

Solution: The numerator and the denominator are both equal to zero at x = 3, but there is a common factor x - 3 that can be removed (that is, we can simplify the rational function):

\begin{align*}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align*} \begin{align*}= \lim_{x \rightarrow 3} \frac{2(x - 3)} {(x + 4) (x - 3)}\end{align*}
\begin{align*}= \lim_{x \rightarrow 3} \frac{2} {x + 4}\end{align*}
\begin{align*}= \frac{2} {7}\end{align*}

## Computing Limits Using One-Sided Limits

When we wish to find the limit of a function f(x) as it approaches a point a and we cannot evaluate f(x) at a because it is undefined at that point, we can compute the function's one-sided limits in order to find the desired limit. If its one-sided limits are the same, then the desired limit exists and is the value of the one-sided limits. If its one-sided limits are not the same, then the desired limit does not exist. This technique is used in the examples below.

Example 8: Find the limit f(x) as x approaches 1. That is, find \begin{align*}\lim_{x \rightarrow 1} f(x)\end{align*} if

\begin{align*}f(x) = \begin{cases}3 - x, & x < 1 \\ 3x - x^2, & x > 1 \end{cases}\end{align*}

Solution: Remember that we are not concerned about finding the value of f(x) at x but rather near x. So, for x < 1 (limit from the left),

\begin{align*}\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (3 - x) = (3 - 1) = 2\end{align*}

and for x > 1 (limit from the right),

\begin{align*}\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x - x^2) = 2\end{align*}

Now since the limit exists and is the same on both sides, it follows that

\begin{align*}\lim_{x \rightarrow 1} f(x) = 2\end{align*}

Example 9: Find \begin{align*}\lim_{x \rightarrow 2} \frac{3} {x - 2}\end{align*}.

Solution: From the figure below we see that \begin{align*}f(x)=\frac{3} {x - 2}\end{align*} decreases without bound as x approaches 2 from the left and \begin{align*}f(x)=\frac{3} {x - 2}\end{align*} increases without bound as x approaches 2 from the right.

This means that \begin{align*}\lim_{x \rightarrow 2^-} \frac{3} {x - 2} = -\infty\end{align*} and \begin{align*}\lim_{x \rightarrow 2^+} \frac{3} {x - 2} = +\infty\end{align*}. Since f(x) is unbounded (infinite) in either directions, the limit does not exist.

## Lesson Summary

In this lesson we learned many techniques for computing limits. In particular, we learned that we can find the limit of a polynomial function approaching a point a by evaluating the function at a. We also learned that the limits of some rational functions do not exist, but sometimes we can simplify the functions in order to compute their limits. Just because a function is discontinuous at a given point does not necessarily mean that it's limit does not exist at that point. Finally, we learned that limits can often be computed by finding both one-sided limits and checking to see if they are equal.

## Points to Consider

1. Can you prove that the theorems in the boxes at the beginning of this lesson are true?
2. Why does it make sense that a limit can exist for a function at a point of discontinuity (as in Example 6)?

## Review Questions

1. Find \begin{align*} \lim_{x \rightarrow -2} (2)\end{align*}.
2. Find \begin{align*} \lim_{x \rightarrow 0^+} (\pi)\end{align*}.
3. Find \begin{align*} \lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2}\end{align*}.
4. Find \begin{align*} \lim_{x \rightarrow 6} \frac{x - 6} {x^2 - 36}\end{align*}.
5. Find \begin{align*} \lim_{x \rightarrow 5} \sqrt{x^3 - 2x - 1}\end{align*}.
6. Find \begin{align*} \lim_{x \rightarrow 3^+} \frac{3} {x - 3}\end{align*}.
7. Show that \begin{align*}\lim_{x \rightarrow 0^+} \left (\frac{1} {x} - \frac{1} {x^2}\right ) = -\infty\end{align*}.
8. For an object in free fall, such as a stone falling off a cliff, the distance y(t) (in meters) that the object falls in t seconds is given by the kinematic equation y(t) = 4.9 t2. The object’s velocity after 2 seconds is given by \begin{align*}v(t) = \lim_{t \rightarrow 2} \frac{y(t) - y(2)} {t - 2}\end{align*}. What is the velocity of the object after 2 seconds?

1. 2
2. \begin{align*}\pi\end{align*}
3. 4
4. \begin{align*}\frac{1} {12}\end{align*}
5. \begin{align*}\sqrt{114}\end{align*}
6. \begin{align*}+\infty\end{align*}
7. Hint: Use a graph.
8. 19.6 m/sec

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