Evaluating Limits with the Limit Laws

The first two limit laws were stated in (Figure) and we repeat them here. These basic results, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

We now practice applying these limit laws to evaluate a limit.

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that [latex]\underset{x\to a}{\lim}f(x)=f(a)[/latex]. This is not always true, but it does hold for all polynomials for any choice of [latex]a[/latex] and for all rational functions at all values of [latex]a[/latex] for which the rational function is defined.

To see that this theorem holds, consider the polynomial [latex]p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0[/latex]. By applying the sum, constant multiple, and power laws, we end up with

It now follows from the quotient law that if [latex]p(x)[/latex] and [latex]q(x)[/latex] are polynomials for which [latex]q(a)\ne 0[/latex], then

(Figure) applies this result.

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 [latex]\underset{x\to a}{\lim}f(x)[/latex] to exist when [latex]f(a)[/latex] is undefined. The following observation allows us to evaluate many limits of this type:

If for all [latex]x\ne a, \, f(x)=g(x)[/latex] over some open interval containing [latex]a[/latex], then [latex]\underset{x\to a}{\lim}f(x)=\underset{x\to a}{\lim}g(x)[/latex].

To understand this idea better, consider the limit [latex]\underset{x\to 1}{\lim}\frac{x^2-1}{x-1}[/latex].

The function

and the function [latex]g(x)=x+1[/latex] are identical for all values of [latex]x\ne 1.[/latex] The graphs of these two functions are shown in (Figure).

Figure 1. The graphs of [latex]f(x)[/latex] and [latex]g(x)[/latex] are identical for all [latex]x\ne 1[/latex]. Their limits at 1 are equal.

We see that

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

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

(Figure) 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.

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 [latex]\underset{x\to a^-}{\lim}h(x)[/latex], we require the function [latex]h(x)[/latex] to be defined over an open interval of the form [latex](b,a)[/latex]; for a limit of the form [latex]\underset{x\to a^+}{\lim}h(x)[/latex], we require the function [latex]h(x)[/latex] to be defined over an open interval of the form [latex](a,c)[/latex]. (Figure) illustrates this point.

Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

1. [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex]
2. [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}[/latex]

Show Solution

(Figure) illustrates the function [latex]f(x)=\sqrt{x-3}[/latex] and aids in our understanding of these limits.

Figure 2. The graph shows the function [latex]f(x)=\sqrt{x-3}[/latex].

1. The function [latex]f(x)=\sqrt{x-3}[/latex] is defined over the interval [latex][3,+\infty)[/latex]. Since this function is not defined to the left of 3, we cannot apply the limit laws to compute [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex]. In fact, since [latex]f(x)=\sqrt{x-3}[/latex] is undefined to the left of 3, [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex] does not exist.
2. Since [latex]f(x)=\sqrt{x-3}[/latex] is defined to the right of 3, the limit laws do apply to [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}[/latex]. By applying these limit laws we obtain [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}=0[/latex].

In (Figure) 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.

Evaluating a Two-Sided Limit Using the Limit Laws

For [latex]f(x)=\begin{cases} 4x-3 & \text{if} \, x<2 \\ (x-3)^2 & \text{if} \, x \ge 2 \end{cases}[/latex] evaluate each of the following limits:

1. [latex]\underset{x\to 2^-}{\lim}f(x)[/latex]
2. [latex]\underset{x\to 2^+}{\lim}f(x)[/latex]
3. [latex]\underset{x\to 2}{\lim}f(x)[/latex]

Show Solution

(Figure) illustrates the function [latex]f(x)[/latex] and aids in our understanding of these limits.

Figure 3. This graph shows the function [latex]f(x)[/latex].

1. Since [latex]f(x)=4x-3[/latex] for all [latex]x[/latex] in [latex](−\infty,2)[/latex], replace [latex]f(x)[/latex] in the limit with [latex]4x-3[/latex] and apply the limit laws:
   [latex]\underset{x\to 2^-}{\lim}f(x)=\underset{x\to 2^-}{\lim}(4x-3)=5[/latex].
2. Since [latex]f(x)=(x-3)^2[/latex] for all [latex]x[/latex] in [latex](2,+\infty)[/latex], replace [latex]f(x)[/latex] in the limit with [latex](x-3)^2[/latex] and apply the limit laws:
   [latex]\underset{x\to 2^+}{\lim}f(x)=\underset{x\to 2^-}{\lim}(x-3)^2=1[/latex].
3. Since [latex]\underset{x\to 2^-}{\lim}f(x)=5[/latex] and [latex]\underset{x\to 2^+}{\lim}f(x)=1[/latex], we conclude that [latex]\underset{x\to 2}{\lim}f(x)[/latex] does not exist.

Graph [latex]f(x)=\begin{cases} -x-2 & \text{if} \, x<-1 \\ 2 & \text{if} \, x = -1 \\ x^3 & \text{if} \, x > -1 \end{cases}[/latex] and evaluate [latex]\underset{x\to -1^-}{\lim}f(x)[/latex].

Show Solution

[latex]\underset{x\to -1^-}{\lim}f(x)=-1[/latex]

Hint

Use the method in (Figure) to evaluate the limit.

We now turn our attention to evaluating a limit of the form [latex]\underset{x\to a}{\lim}\large \frac{f(x)}{g(x)}[/latex], where [latex]\underset{x\to a}{\lim}f(x)=K[/latex], where [latex]K\ne 0[/latex] and [latex]\underset{x\to a}{\lim}g(x)=0[/latex]. That is, [latex]f(x)/g(x)[/latex] has the form [latex]K/0, \, K\ne 0[/latex] at [latex]a[/latex].

The Squeeze Theorem

The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point [latex]a[/latex] that is unknown, between two functions having a common known limit at [latex]a[/latex]. (Figure) illustrates this idea.

Figure 4. The Squeeze Theorem applies when [latex]f(x)\le g(x)\le h(x)[/latex] and [latex]\underset{x\to a}{\lim}f(x)=\underset{x\to a}{\lim}h(x)[/latex].

Applying the Squeeze Theorem

Apply the Squeeze Theorem to evaluate [latex]\underset{x\to 0}{\lim}x \cos x[/latex].

Show Solution

Because [latex]-1\le \cos x\le 1[/latex] for all [latex]x[/latex], we have [latex]-x\le x \cos x\le x[/latex] for [latex]x\ge 0[/latex] and [latex]-x\ge xcosx\ge x[/latex] for [latex]x\le 0[/latex] (if [latex]x[/latex] is negative the direction of the inequalities changes when we multiply). Since [latex]\underset{x\to 0}{\lim}(-x)=0=\underset{x\to 0}{\lim}x[/latex], from the Squeeze Theorem we obtain [latex]\underset{x\to 0}{\lim}x \cos x=0[/latex]. The graphs of [latex]f(x)=-x, \, g(x)=x \cos x[/latex], and [latex]h(x)=x[/latex] are shown in (Figure).

Figure 5. The graphs of [latex]f(x), \, g(x)[/latex], and [latex]h(x)[/latex] are shown around the point [latex]x=0[/latex].

We now use the Squeeze Theorem to tackle several very important limits. Although this discussion is somewhat lengthy, these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of these limits is [latex]\underset{\theta \to 0}{\lim} \sin \theta[/latex]. Consider the unit circle shown in (Figure). In the figure, we see that [latex]\sin \theta[/latex] is the [latex]y[/latex]-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for [latex]0<\theta <\frac{\pi }{2}, \, 0 < \sin \theta < \theta[/latex].

Figure 6. The sine function is shown as a line on the unit circle.

Because [latex]\underset{\theta \to 0^+}{\lim}0=0[/latex] and [latex]\underset{\theta \to 0^+}{\lim}\theta =0[/latex], by using the Squeeze Theorem we conclude that

To see that [latex]\underset{\theta \to 0^-}{\lim} \sin \theta =0[/latex] as well, observe that for [latex]-\frac{\pi }{2} < \theta <0, \, 0 < −\theta < \frac{\pi}{2}[/latex] and hence, [latex]0 < \sin(-\theta) < −\theta[/latex]. Consequently, [latex]0 < -\sin \theta < −\theta[/latex] It follows that [latex]0 > \sin \theta > \theta[/latex]. An application of the Squeeze Theorem produces the desired limit. Thus, since [latex]\underset{\theta \to 0^+}{\lim} \sin \theta =0[/latex] and [latex]\underset{\theta \to 0^-}{\lim} \sin \theta =0[/latex],

Next, using the identity [latex]\cos \theta =\sqrt{1-\sin^2 \theta}[/latex] for [latex]-\frac{\pi}{2}<\theta <\frac{\pi}{2}[/latex], we see that

We now take a look at a limit that plays an important role in later chapters—namely, [latex]\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}[/latex]. To evaluate this limit, we use the unit circle in (Figure). Notice that this figure adds one additional triangle to (Figure). We see that the length of the side opposite angle [latex]\theta[/latex] in this new triangle is [latex]\tan \theta[/latex]. Thus, we see that for [latex]0 < \theta < \frac{\pi}{2}, \, \sin \theta < \theta < \tan \theta[/latex].

Figure 7. The sine and tangent functions are shown as lines on the unit circle.

By dividing by [latex]\sin \theta[/latex] in all parts of the inequality, we obtain

Equivalently, we have

Since [latex]\underset{\theta \to 0^+}{\lim}1=1=\underset{\theta \to 0^+}{\lim}\cos \theta[/latex], we conclude that [latex]\underset{\theta \to 0^+}{\lim}\frac{\sin \theta}{\theta}=1[/latex]. By applying a manipulation similar to that used in demonstrating that [latex]\underset{\theta \to 0^-}{\lim}\sin \theta =0[/latex], we can show that [latex]\underset{\theta \to 0^-}{\lim}\frac{\sin \theta}{\theta}=1[/latex]. Thus,

In (Figure) we use this limit to establish [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0[/latex]. This limit also proves useful in later chapters.