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Additional Limit Evaluation Techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right)\) to exist when \(f \left(\right. a \left.\right)\) is undefined. The following observation allows us to evaluate many limits of this type:

If for all \(x \neq a , f \left(\right. x \left.\right) = g \left(\right. x \left.\right)\) over some open interval containing a, then \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = \underset{x \rightarrow a}{\text{lim}} g \left(\right. x \left.\right) .\)

To understand this idea better, consider the limit \(\underset{x \rightarrow 1}{\text{lim}} \frac{x^{2} - 1}{x - 1} .\)

The function

\[\begin{aligned} f \left(\right. x \left.\right) & = \frac{x^{2} - 1}{x - 1} \\ & = \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x - 1} \end{aligned}\]

and the function \(g \left(\right. x \left.\right) = x + 1\) are identical for all values of \(x \neq 1 .\) The graphs of these two functions are shown in Figure 2.24.

Two graphs side by side. The first is a graph of g(x) = x + 1, a linear function with y intercept at (0,1) and x intercept at (-1,0). The second is a graph of f(x) = (x^2 – 1) / (x – 1). This graph is identical to the first for all x not equal to 1, as there is an open circle at (1,2) in the second graph.
Figure 2.24 The graphs of \(f \left(\right. x \left.\right)\) and \(g \left(\right. x \left.\right)\) are identical for all \(x \neq 1 .\) Their limits at 1 are equal.

We see that

\[\begin{aligned} \underset{x \rightarrow 1}{\text{lim}} \frac{x^{2} - 1}{x - 1} & = \underset{x \rightarrow 1}{\text{lim}} \frac{\left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)}{x - 1} \\ & = \underset{x \rightarrow 1}{\text{lim}} \left(\right. x + 1 \left.\right) \\ & = 2 . \end{aligned}\]

The limit has the form \(\underset{x \rightarrow a}{\text{lim}} \frac{f \left(\right. x \left.\right)}{g \left(\right. x \left.\right)} ,\) where \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = 0\) and \(\underset{x \rightarrow a}{\text{lim}} g \left(\right. x \left.\right) = 0 .\) (In this case, we say that \(f \left(\right. x \left.\right) / g \left(\right. x \left.\right)\) has the indeterminate form \(0 / 0 .)\) The following Problem-Solving Strategy provides a general outline for evaluating limits of this type.

Problem-Solving Strategy

Calculating a Limit When \(f \left(\right. x \left.\right) / g \left(\right. x \left.\right)\) has the Indeterminate Form 0/0

  1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately using the limit laws.
  2. We then need to find a function that is equal to \(h \left(\right. x \left.\right) = f \left(\right. x \left.\right) / g \left(\right. x \left.\right)\) for all \(x \neq a\) over some interval containing a. To do this, we may need to try one or more of the following steps:
    1. If \(f \left(\right. x \left.\right)\) and \(g \left(\right. x \left.\right)\) are polynomials, we should factor each function and cancel out any common factors.
    2. If the numerator or denominator contains a difference involving a square root, we should try multiplying the numerator and denominator by the conjugate of the expression involving the square root.
    3. If \(f \left(\right. x \left.\right) / g \left(\right. x \left.\right)\) is a complex fraction, we begin by simplifying it.
  3. Last, we apply the limit laws.

The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19, we look at simplifying a complex fraction.

Example 2.17

Evaluating a Limit by Factoring and Canceling

Evaluate \(\underset{x \rightarrow 3}{\text{lim}} \frac{x^{2} - 3 x}{2 x^{2} - 5 x - 3} .\)

Solution

Step 1. The function \(f \left(\right. x \left.\right) = \frac{x^{2} - 3 x}{2 x^{2} - 5 x - 3}\) is undefined for \(x = 3 .\) In fact, if we substitute 3 into the function we get \(0 / 0 ,\) which is indeterminate. Factoring and canceling is a good strategy:

\[\underset{x \rightarrow 3}{\text{lim}} \frac{x^{2} - 3 x}{2 x^{2} - 5 x - 3} = \underset{x \rightarrow 3}{\text{lim}} \frac{x \left(\right. x - 3 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. 2 x + 1 \left.\right)}\]

Step 2. For all \(x \neq 3 , \frac{x^{2} - 3 x}{2 x^{2} - 5 x - 3} = \frac{x}{2 x + 1} .\) Therefore,

\[\underset{x \rightarrow 3}{\text{lim}} \frac{x \left(\right. x - 3 \left.\right)}{\left(\right. x - 3 \left.\right) \left(\right. 2 x + 1 \left.\right)} = \underset{x \rightarrow 3}{\text{lim}} \frac{x}{2 x + 1} .\]

Step 3. Evaluate using the limit laws:

\[\underset{x \rightarrow 3}{\text{lim}} \frac{x}{2 x + 1} = \frac{3}{7} .\]

Checkpoint 2.13

Evaluate \(\underset{x \rightarrow −3}{\text{lim}} \frac{x^{2} + 4 x + 3}{x^{2} - 9} .\)

Example 2.18

Evaluating a Limit by Multiplying by a Conjugate

Evaluate \(\underset{x \rightarrow −1}{\text{lim}} \frac{\sqrt{x + 2} - 1}{x + 1} .\)

Solution

Step 1. \(\frac{\sqrt{x + 2} - 1}{x + 1}\) has the form \(0 / 0\) at −1. Let’s begin by multiplying by \(\sqrt{x + 2} + 1 ,\) the conjugate of \(\sqrt{x + 2} - 1 ,\) on the numerator and denominator:

\[\underset{x \rightarrow −1}{\text{lim}} \frac{\sqrt{x + 2} - 1}{x + 1} = \underset{x \rightarrow −1}{\text{lim}} \frac{\sqrt{x + 2} - 1}{x + 1} \cdot \frac{\sqrt{x + 2} + 1}{\sqrt{x + 2} + 1} .\]

Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that the \(\left(\right. x + 1 \left.\right)\) in the denominator cancels out in the end:

\[= \underset{x \rightarrow −1}{\text{lim}} \frac{x + 1}{\left(\right. x + 1 \left.\right) \left(\right. \sqrt{x + 2} + 1 \left.\right)} .\]

Step 3. Then we cancel:

\[= \underset{x \rightarrow −1}{\text{lim}} \frac{1}{\sqrt{x + 2} + 1} .\]

Step 4. Last, we apply the limit laws:

\[\underset{x \rightarrow −1}{\text{lim}} \frac{1}{\sqrt{x + 2} + 1} = \frac{1}{2} .\]

Checkpoint 2.14

Evaluate \(\underset{x \rightarrow 5}{\text{lim}} \frac{\sqrt{x - 1} - 2}{x - 5} .\)

Example 2.19

Evaluating a Limit by Simplifying a Complex Fraction

Evaluate \(\underset{x \rightarrow 1}{\text{lim}} \frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1} .\)

Solution

Step 1. \(\frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1}\) has the form \(0 / 0\) at 1. We simplify the algebraic fraction by multiplying by \(2 \left(\right. x + 1 \left.\right) / 2 \left(\right. x + 1 \left.\right) :\)

\[\underset{x \rightarrow 1}{\text{lim}} \frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1} = \underset{x \rightarrow 1}{\text{lim}} \frac{\frac{1}{x + 1} - \frac{1}{2}}{x - 1} \cdot \frac{2 \left(\right. x + 1 \left.\right)}{2 \left(\right. x + 1 \left.\right)} .\]

Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able to cancel the factor \(\left(\right. x - 1 \left.\right) :\)

\[= \underset{x \rightarrow 1}{\text{lim}} \frac{2 - \left(\right. x + 1 \left.\right)}{2 \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} .\]

Step 3. Then, we simplify the numerator:

\[= \underset{x \rightarrow 1}{\text{lim}} \frac{- x + 1}{2 \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} .\]

Step 4. Now we factor out −1 from the numerator:

\[= \underset{x \rightarrow 1}{\text{lim}} \frac{- \left(\right. x - 1 \left.\right)}{2 \left(\right. x - 1 \left.\right) \left(\right. x + 1 \left.\right)} .\]

Step 5. Then, we cancel the common factors of \(\left(\right. x - 1 \left.\right) :\)

\[= \underset{x \rightarrow 1}{\text{lim}} \frac{−1}{2 \left(\right. x + 1 \left.\right)} .\]

Step 6. Last, we evaluate using the limit laws:

\[\underset{x \rightarrow 1}{\text{lim}} \frac{−1}{2 \left(\right. x + 1 \left.\right)} = - \frac{1}{4} .\]

Checkpoint 2.15

Evaluate \(\underset{x \rightarrow −3}{\text{lim}} \frac{\frac{1}{x + 2} + 1}{x + 3} .\)

Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little creativity, we can still use these same techniques.

Example 2.20

Evaluating a Limit When the Limit Laws Do Not Apply

Evaluate \(\underset{x \rightarrow 0}{\text{lim}} \left(\right. \frac{1}{x} + \frac{5}{x \left(\right. x - 5 \left.\right)} \left.\right) .\)

Solution

Both \(1 / x\) and \(5 / x \left(\right. x - 5 \left.\right)\) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing addition and then applying one of our previous strategies. Observe that

\[\begin{aligned} \frac{1}{x} + \frac{5}{x \left(\right. x - 5 \left.\right)} & = \frac{x - 5 + 5}{x \left(\right. x - 5 \left.\right)} \\ & = \frac{x}{x \left(\right. x - 5 \left.\right)} . \end{aligned}\]

Thus,

\[\begin{aligned} \underset{x \rightarrow 0}{\text{lim}} \left(\right. \frac{1}{x} + \frac{5}{x \left(\right. x - 5 \left.\right)} \left.\right) & = \underset{x \rightarrow 0}{\text{lim}} \frac{x}{x \left(\right. x - 5 \left.\right)} \\ & = \underset{x \rightarrow 0}{\text{lim}} \frac{1}{x - 5} \\ & = - \frac{1}{5} . \end{aligned}\]

Checkpoint 2.16

Evaluate \(\underset{x \rightarrow 3}{\text{lim}} \left(\right. \frac{1}{x - 3} - \frac{4}{x^{2} - 2 x - 3} \left.\right) .\)

Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form \(\underset{x \rightarrow a^{-}}{\text{lim}} h \left(\right. x \left.\right) ,\) we require the function \(h \left(\right. x \left.\right)\) to be defined over an open interval of the form \(\left(\right. b , a \left.\right) ;\) for a limit of the form \(\underset{x \rightarrow a^{+}}{\text{lim}} h \left(\right. x \left.\right) ,\) we require the function \(h \left(\right. x \left.\right)\) to be defined over an open interval of the form \(\left(\right. a , c \left.\right) .\) Example 2.21 illustrates this point.

Example 2.21

Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

  1. \(\underset{x \rightarrow 3^{-}}{\text{lim}} \sqrt{x - 3}\)
  2. \(\underset{x \rightarrow 3^{+}}{\text{lim}} \sqrt{x - 3}\)

Solution

Figure 2.25 illustrates the function \(f \left(\right. x \left.\right) = \sqrt{x - 3}\) and aids in our understanding of these limits.

A graph of the function f(x) = sqrt(x-3). Visually, the function looks like the top half of a parabola opening to the right with vertex at (3,0).
Figure 2.25 The graph shows the function \(f \left(\right. x \left.\right) = \sqrt{x - 3} .\)
  1. The function \(f \left(\right. x \left.\right) = \sqrt{x - 3}\) is defined over the interval \(\left[\right. 3 , + \infty \left.\right) .\) Since this function is not defined to the left of 3, we cannot apply the limit laws to compute \(\underset{x \rightarrow 3^{-}}{\text{lim}} \sqrt{x - 3} .\) In fact, since \(f \left(\right. x \left.\right) = \sqrt{x - 3}\) is undefined to the left of 3, \(\underset{x \rightarrow 3^{-}}{\text{lim}} \sqrt{x - 3}\) does not exist.
  2. Since \(f \left(\right. x \left.\right) = \sqrt{x - 3}\) is defined to the right of 3, the limit laws do apply to \(\underset{x \rightarrow 3^{+}}{\text{lim}} \sqrt{x - 3} .\) By applying these limit laws we obtain \(\underset{x \rightarrow 3^{+}}{\text{lim}} \sqrt{x - 3} = 0 .\)

In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Example 2.22

Evaluating a Two-Sided Limit Using the Limit Laws

For \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. 4 x - 3 & \text{if} x < 2 \\ \left(\right. x - 3 \left.\right)^{2} & \text{if} x \geq 2 , \end{aligned}\) evaluate each of the following limits:

  1. \(\underset{x \rightarrow 2^{-}}{\text{lim}} f \left(\right. x \left.\right)\)
  2. \(\underset{x \rightarrow 2^{+}}{\text{lim}} f \left(\right. x \left.\right)\)
  3. \(\underset{x \rightarrow 2}{\text{lim}} f \left(\right. x \left.\right)\)

Solution

Figure 2.26 illustrates the function \(f \left(\right. x \left.\right)\) and aids in our understanding of these limits.

The graph of a piecewise function with two segments. For x<2, the function is linear with the equation 4x-3. There is an open circle at (2,5). The second segment is a parabola and exists for x>=2, with the equation (x-3)^2. There is a closed circle at (2,1). The vertex of the parabola is at (3,0).
Figure 2.26 This graph shows a function \(f \left(\right. x \left.\right) .\)
  1. Since \(f \left(\right. x \left.\right) = 4 x - 3\) for all x in \(\left(\right. − \infty , 2 \left.\right) ,\) replace \(f \left(\right. x \left.\right)\) in the limit with \(4 x - 3\) and apply the limit laws:
    \[\underset{x \rightarrow 2^{-}}{\text{lim}} f \left(\right. x \left.\right) = \underset{x \rightarrow 2^{-}}{\text{lim}} \left(\right. 4 x - 3 \left.\right) = 5 .\]
  2. Since \(f \left(\right. x \left.\right) = \left(\right. x - 3 \left.\right)^{2}\) for all x in \(\left(\right. 2 , + \infty \left.\right) ,\) replace \(f \left(\right. x \left.\right)\) in the limit with \(\left(\right. x - 3 \left.\right)^{2}\) and apply the limit laws:
    \[\underset{x \rightarrow 2^{+}}{\text{lim}} f \left(\right. x \left.\right) = \underset{x \rightarrow 2^{+}}{\text{lim}} \left(\right. x - 3 \left.\right)^{2} = 1 .\]
  3. Since \(\underset{x \rightarrow 2^{-}}{\text{lim}} f \left(\right. x \left.\right) = 5\) and \(\underset{x \rightarrow 2^{+}}{\text{lim}} f \left(\right. x \left.\right) = 1 ,\) we conclude that \(\underset{x \rightarrow 2}{\text{lim}} f \left(\right. x \left.\right)\) does not exist.

Checkpoint 2.17

Graph \(\begin{aligned} f \left(\right. x \left.\right) = \left{\right. - x - 2 \text{if} x < − 1 \\ 2 \text{if} x = −1 \\ x^{3} \text{if} x > − 1 \end{aligned}\) and evaluate \(\underset{x \rightarrow −1^{-}}{\text{lim}} f \left(\right. x \left.\right) .\)

We now turn our attention to evaluating a limit of the form \(\underset{x \rightarrow a}{\text{lim}} \frac{f \left(\right. x \left.\right)}{g \left(\right. x \left.\right)} ,\) where \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = K ,\) where \(K \neq 0\) and \(\underset{x \rightarrow a}{\text{lim}} g \left(\right. x \left.\right) = 0 .\) That is, \(f \left(\right. x \left.\right) / g \left(\right. x \left.\right)\) has the form \(K / 0 , K \neq 0\) at a.

Example 2.23

Evaluating a Limit of the Form \(K / 0 , K \neq 0\) Using the Limit Laws

Evaluate \(\underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x^{2} - 2 x} .\)

Solution

Step 1. After substituting in \(x = 2 ,\) we see that this limit has the form \(−1 / 0 .\) That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of \(\frac{x - 3}{x \left(\right. x - 2 \left.\right)}\) becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

\[\underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x^{2} - 2 x} = \underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x \left(\right. x - 2 \left.\right)} .\]

Step 2. Since \(x - 2\) is the only part of the denominator that is zero when 2 is substituted, we then separate \(1 / \left(\right. x - 2 \left.\right)\) from the rest of the function:

\[= \underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x} \cdot \frac{1}{x - 2} .\]

Step 3. \(\underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x} = - \frac{1}{2}\) and \(\underset{x \rightarrow 2^{-}}{\text{lim}} \frac{1}{x - 2} = − \infty .\) Therefore, the product of \(\left(\right. x - 3 \left.\right) / x\) and \(1 / \left(\right. x - 2 \left.\right)\) has a limit of \(+\infty:\)

\[\underset{x \rightarrow 2^{-}}{\text{lim}} \frac{x - 3}{x^{2} - 2 x} = + \infty .\]

Checkpoint 2.18

Evaluate \(\underset{x \rightarrow 1}{\text{lim}} \frac{x + 2}{\left(\right. x - 1 \left.\right)^{2}} .\)

This lesson is part of:

Limits

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