If F(X) Is a Polynomial Then Lim F(X)=f(C) True/false Homework Exercises Review

2. Limits

2.3 The Limit Laws

Learning Objectives

Recognize the basic limit laws.
Use the limit laws to evaluate the limit of a function.
Evaluate the limit of a function by factoring.
Use the limit laws to evaluate the limit of a polynomial or rational function.
Evaluate the limit of a role by factoring or by using conjugates.
Evaluate the limit of a part by using the squeeze theorem.

In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we establish laws for calculating limits and larn how to apply these laws. In the Educatee Projection at the finish of this department, you have the opportunity to apply these limit laws to derive the formula for the surface area of a circle by adapting a method devised past the Greek mathematician Archimedes. We begin by restating 2 useful limit results from the previous department. These ii results, together with the limit laws, serve every bit a foundation for computing many limits.

Evaluating Limits with the Limit Laws

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

Basic Limit Results

For any existent number $a$ and any abiding $c$ ,

$\underset{x\to a}{\lim}x=a$
$\underset{x\to a}{\lim}c=c$

Evaluating a Basic Limit

Evaluate each of the following limits using (Figure).

$\underset{x\to 2}{\lim}x$
$\underset{x\to 2}{\lim}5$

We at present take a look at the limit laws, the individual backdrop of limits. The proofs that these laws hold are omitted hither.

We at present do applying these limit laws to evaluate a limit.

Evaluating a Limit Using Limit Laws

Employ the limit laws to evaluate $\underset{x\to -3}{\lim}(4x+2)$ .

Solution

Let'southward apply the limit laws one step at a time to be sure we understand how they work. We need to proceed in mind the requirement that, at each application of a limit police force, the new limits must exist for the limit law to exist applied.

$\begin{array}{ccccc}\underset{x\to -3}{\lim}(4x+2)\hfill & =\underset{x\to -3}{\lim}4x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the sum law.}\hfill \\ & =4 \cdot \underset{x\to -3}{\lim}x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the constant multiple law.}\hfill \\ & =4 \cdot (-3)+2=-10\hfill & & & \text{Apply the basic limit results and simplify.}\hfill \end{array}$

Using Limit Laws Repeatedly

Use the limit laws to evaluate $\underset{x\to 2}{\lim}\frac{2x^2-3x+1}{x^3+4}$ .

Solution

To find this limit, we need to apply the limit laws several times. Over again, nosotros demand to go on in heed that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to exist applied.

$\begin{array}{ccccc}\\ \\ \underset{x\to 2}{\lim}\large \frac{2x^2-3x+1}{x^3+4} & = \large \frac{\underset{x\to 2}{\lim}(2x^2-3x+1)}{\underset{x\to 2}{\lim}(x^3+4)} & & & \text{Apply the quotient law, making sure that} \, 2^3+4\ne 0 \\ & = \large \frac{2 \cdot \underset{x\to 2}{\lim}x^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{\underset{x\to 2}{\lim}x^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the sum law and constant multiple law.} \\ & = \large \frac{2 \cdot (\underset{x\to 2}{\lim}x)^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{(\underset{x\to 2}{\lim}x)^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the power law.} \\ & = \large \frac{2(4)-3(2)+1}{2^3+4}=\frac{1}{4} & & & \text{Apply the basic limit laws and simplify.} \end{array}$

Utilize the limit laws to evaluate $\underset{x\to 6}{\lim}(2x-1)\sqrt{x+4}$ . In each step, indicate the limit law applied.

Solution

$11\sqrt{10}$

Limits of Polynomial and Rational Functions

By now y'all accept probably noticed that, in each of the previous examples, it has been the instance that $\underset{x\to a}{\lim}f(x)=f(a)$ . This is not e'er true, but information technology does hold for all polynomials for any choice of $a$ and for all rational functions at all values of $a$ for which the rational function is defined.

Limits of Polynomial and Rational Functions

Let $p(x)$ and $q(x)$ be polynomial functions. Permit $a$ be a real number. And so,

$\underset{x\to a}{\lim}p(x)=p(a)$

$\underset{x\to a}{\lim}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)} \, \text{when} \, q(a)\ne 0$ .

To see that this theorem holds, consider the polynomial $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$ . By applying the sum, constant multiple, and ability laws, nosotros cease up with

$\begin{array}{cc}\hfill \underset{x\to a}{\lim}p(x)& =\underset{x\to a}{\lim}(c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0)\hfill \\ & =c_n(\underset{x\to a}{\lim}x)^n+c_{n-1}(\underset{x\to a}{\lim}x)^{n-1}+\cdots +c_1(\underset{x\to a}{\lim}x)+\underset{x\to a}{\lim}c_0\hfill \\ & =c_na^n+c_{n-1}a^{n-1}+\cdots +c_1a+c_0\hfill \\ & =p(a)\hfill \end{array}$

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

$\underset{x\to a}{\lim}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}$ .

(Figure) applies this effect.

Evaluating a Limit of a Rational Function

Evaluate the $\underset{x\to 3}{\lim}\frac{2x^2-3x+1}{5x+4}$ .

Solution

Since iii is in the domain of the rational role $f(x)=\frac{2x^2-3x+1}{5x+4}$ , nosotros tin calculate the limit by substituting 3 for $x$ into the function. Thus,

$\underset{x\to 3}{\lim}\frac{2x^2-3x+1}{5x+4}=\frac{10}{19}$ .

Evaluate $\underset{x\to -2}{\lim}(3x^3-2x+7)$ .

Additional Limit Evaluation Techniques

As we have seen, we may evaluate easily the limits of polynomials and limits of some (only not all) rational functions by direct substitution. Withal, equally nosotros saw in the introductory section on limits, it is certainly possible for $\underset{x\to a}{\lim}f(x)$ to exist when $f(a)$ is undefined. The following observation allows united states to evaluate many limits of this type:

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

To understand this thought amend, consider the limit $\underset{x\to 1}{\lim}\frac{x^2-1}{x-1}$ .

The function

$\begin{array}{cc}\hfill f(x)& =\frac{x^2-1}{x-1}\hfill \\ & =\frac{(x-1)(x+1)}{x-1}\hfill \end{array}$

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

We see that

$\begin{array}{cc}\hfill \underset{x\to 1}{\lim}\frac{x^2-1}{x-1}& =\underset{x\to 1}{\lim}\frac{(x-1)(x+1)}{x-1}\hfill \\ & =\underset{x\to 1}{\lim}(x+1)\hfill \\ & =2\hfill \end{array}$

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

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

Evaluating a Limit by Factoring and Canceling

Evaluate $\underset{x\to 3}{\lim}\frac{x^2-3x}{2x^2-5x-3}$ .

Solution

Step 1. The function $f(x)=\frac{x^2-3x}{2x^2-5x-3}$ is undefined for $x=3$ . In fact, if we substitute iii into the office we become 0/0, which is undefined. Factoring and canceling is a skilful strategy:

$\underset{x\to 3}{\lim}\frac{x^2-3x}{2x^2-5x-3}=\underset{x\to 3}{\lim}\frac{x(x-3)}{(x-3)(2x+1)}$

Stride 2. For all $x\ne 3, \, \frac{x^2-3x}{2x^2-5x-3}=\frac{x}{2x+1}$ . Therefore,

$\underset{x\to 3}{\lim}\frac{x(x-3)}{(x-3)(2x+1)}=\underset{x\to 3}{\lim}\frac{x}{2x+1}$ .

Footstep three. Evaluate using the limit laws:

$\underset{x\to 3}{\lim}\frac{x}{2x+1}=\frac{3}{7}$ .

Evaluate $\underset{x\to -3}{\lim}\frac{x^2+4x+3}{x^2-9}$ .

Solution

$\frac{1}{3}$

Evaluating a Limit by Multiplying by a Conjugate

Evaluate $\underset{x\to -1}{\lim}\frac{\sqrt{x+2}-1}{x+1}$ .

Solution

Stride ane. $\frac{\sqrt{x+2}-1}{x+1}$ 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:

$\underset{x\to -1}{\lim}\frac{\sqrt{x+2}-1}{x+1}=\underset{x\to -1}{\lim}\frac{\sqrt{x+2}-1}{x+1}\cdot \frac{\sqrt{x+2}+1}{\sqrt{x+2}+1}$ .

Footstep two. Nosotros then multiply out the numerator. We don't multiply out the denominator considering we are hoping that the $(x+1)$ in the denominator cancels out in the end:

$=\underset{x\to -1}{\lim}\frac{x+1}{(x+1)(\sqrt{x+2}+1)}$ .

Step three. And then we cancel:

$=\underset{x\to -1}{\lim}\frac{1}{\sqrt{x+2}+1}$ .

Step 4. Concluding, we utilize the limit laws:

$\underset{x\to -1}{\lim}\frac{1}{\sqrt{x+2}+1}=\frac{1}{2}$ .

Evaluate $\underset{x\to 5}{\lim}\frac{\sqrt{x-1}-2}{x-5}$ .

Solution

$\frac{1}{4}$

Evaluating a Limit by Simplifying a Complex Fraction

Evaluate $\underset{x\to 1}{\lim}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}$ .

Solution

Step 1. $\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}$ has the form 0/0 at i. Nosotros simplify the algebraic fraction by multiplying past $2(x+1)/2(x+1)$ :

$\underset{x\to 1}{\lim}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1}=\underset{x\to 1}{\lim}\frac{\frac{1}{x+1}-\frac{1}{2}}{x-1} \cdot \frac{2(x+1)}{2(x+1)}$ .

Pace 2. Next, we multiply through the numerators. Practice not multiply the denominators considering nosotros desire to be able to cancel the factor $(x-1)$ :

$=\underset{x\to 1}{\lim}\frac{2-(x+1)}{2(x-1)(x+1)}$ .

Pace iii. Then, nosotros simplify the numerator:

$=\underset{x\to 1}{\lim}\frac{-x+1}{2(x-1)(x+1)}$ .

Step four. Now nosotros factor out −one from the numerator:

$=\underset{x\to 1}{\lim}\frac{-(x-1)}{2(x-1)(x+1)}$ .

Pace 5. Then, we abolish the mutual factors of $(x-1)$ :

$=\underset{x\to 1}{\lim}\frac{-1}{2(x+1)}$ .

Pace 6. Last, nosotros evaluate using the limit laws:

$\underset{x\to 1}{\lim}\frac{-1}{2(x+1)}=-\frac{1}{4}$ .

Evaluate $\underset{x\to -3}{\lim}\frac{\frac{1}{x+2}+1}{x+3}$ .

(Figure) does non fall neatly into any of the patterns established in the previous examples. However, with a little inventiveness, we can still use these same techniques.

Evaluating a Limit When the Limit Laws Do Non Apply

Evaluate $\underset{x\to 0}{\lim}\big(\frac{1}{x}+\frac{5}{x(x-5)}\big)$ .

Solution

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

$\begin{array}{cc} \frac{1}{x}+\frac{5}{x(x-5)}& =\frac{x-5+5}{x(x-5)} \\ & =\frac{x}{x(x-5)}\end{array}$

Thus,

$\begin{array}{cc}\underset{x\to 0}{\lim}\big(\frac{1}{x}+\frac{5}{x(x-5)}\big)& =\underset{x\to 0}{\lim}\frac{x}{x(x-5)} \\ & =\underset{x\to 0}{\lim}\frac{1}{x-5} \\ & =-\frac{1}{5} \end{array}$

Evaluate $\underset{x\to 3}{\lim}(\frac{1}{x-3}-\frac{4}{x^2-2x-3})$ .

Solution

$\frac{1}{4}$

Let'due south at present revisit one-sided limits. Simple modifications in the limit laws permit the states to apply them to 1-sided limits. For example, to apply the limit laws to a limit of the class $\underset{x\to a^-}{\lim}h(x)$ , we crave the function $h(x)$ to be divers over an open interval of the form $(b,a)$ ; for a limit of the form $\underset{x\to a^+}{\lim}h(x)$ , we require the function $h(x)$ to be defined over an open interval of the form $(a,c)$ . (Figure) illustrates this point.

Evaluating a 1-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

$\underset{x\to 3^-}{\lim}\sqrt{x-3}$
$\underset{x\to 3^+}{\lim}\sqrt{x-3}$

In (Figure) we look at one-sided limits of a piecewise-defined function and use these limits to depict a conclusion nigh a ii-sided limit of the same function.

Evaluating a Two-Sided Limit Using the Limit Laws

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

Solution

-1, which crossed the x axis and y centrality at the origin.">
$\underset{x\to -1^-}{\lim}f(x)=-1$

We at present plough our attending to evaluating a limit of the grade $\underset{x\to a}{\lim}\large \frac{f(x)}{g(x)}$ , where $\underset{x\to a}{\lim}f(x)=K$ , where $K\ne 0$ and $\underset{x\to a}{\lim}g(x)=0$ . That is, $f(x)/g(x)$ has the form $K/0, \, K\ne 0$ at $a$ .

Evaluating a Limit of the Form $K/0, \, K\ne 0$ Using the Limit Laws

Evaluate $\underset{x\to 2^-}{\lim}\frac{x-3}{x^2-2x}$ .

Evaluate $\underset{x\to 1}{\lim}\frac{x+2}{(x-1)^2}$ .

Solution

$+\infty$

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 u.s. to calculate limits by "squeezing" a function, with a limit at a point $a$ that is unknown, between two functions having a common known limit at $a$ . (Figure) illustrates this idea.

The Squeeze Theorem

Let $f(x), \, g(x)$ , and $h(x)$ be defined for all $x\ne a$ over an open interval containing $a$ . If

$f(x)\le g(x)\le h(x)$

for all $x\ne a$ in an open interval containing $a$ and

$\underset{x\to a}{\lim}f(x)=L=\underset{x\to a}{\lim}h(x)$

where $L$ is a real number, then $\underset{x\to a}{\lim}g(x)=L$ .

Applying the Squeeze Theorem

Utilize the Squeeze Theorem to evaluate $\underset{x\to 0}{\lim}x \cos x$ .

Use the Squeeze Theorem to evaluate $\underset{x\to 0}{\lim}x^2 \sin \frac{1}{x}$ .

We now use the Clasp Theorem to tackle several very important limits. Although this give-and-take is somewhat lengthy, these limits show invaluable for the development of the material in both the next section and the side by side chapter. The beginning of these limits is $\underset{\theta \to 0}{\lim} \sin \theta$ . Consider the unit circle shown in (Figure). In the figure, we come across that $\sin \theta$ is the $y$ -coordinate on the unit circle and it corresponds to the line segment shown in blueish. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for $0<\theta <\frac{\pi }{2}, \, 0 < \sin \theta < \theta$ .

A diagram of the unit of measurement circumvolve in the 10,y plane – it is a circle with radius ane and center at the origin. A specific point (cos(theta), sin(theta)) is labeled in quadrant 1 on the edge of the circle. This point is one vertex of a right triangle inside the circle, with other vertices at the origin and (cos(theta), 0). Every bit such, the lengths of the sides are cos(theta) for the base and sin(theta) for the peak, where theta is the angle created by the hypotenuse and base. The radian measure out of angle theta is the length of the arc it subtends on the unit of measurement circle. The diagram shows that for 0 < theta < pi/ii, 0 < sin(theta) < theta. — Effigy 6. The sine function is shown every bit a line on the unit of measurement circle.

Considering $\underset{\theta \to 0^+}{\lim}0=0$ and $\underset{\theta \to 0^+}{\lim}\theta =0$ , by using the Clasp Theorem we conclude that

$\underset{\theta \to 0^+}{\lim} \sin \theta =0$ .

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

$\underset{\theta \to 0}{\lim} \sin \theta =0$ .

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

$\underset{\theta \to 0}{\lim} \cos \theta =\underset{\theta \to 0}{\lim}\sqrt{1-\sin^2 \theta }=1$ .

Nosotros at present take a wait at a limit that plays an of import function in later chapters—namely, $\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}$ . To evaluate this limit, we use the unit of measurement circle in (Figure). Detect that this effigy adds i additional triangle to (Figure). Nosotros see that the length of the side opposite angle $\theta$ in this new triangle is $\tan \theta$ . Thus, we run into that for $0 < \theta < \frac{\pi}{2}, \, \sin \theta < \theta < \tan \theta$ .

The same diagram as the previous ane. Still, the triangle is expanded. The base is now from the origin to (i,0). The height goes from (1,0) to (1, tan(theta)). The hypotenuse goes from the origin to (1, tan(theta)). As such, the height is now tan(theta). Information technology shows that for 0 < theta < pi/2, sin(theta) < theta < tan(theta). — Effigy 7. The sine and tangent functions are shown as lines on the unit of measurement circumvolve.

Past dividing by $\sin \theta$ in all parts of the inequality, nosotros obtain

$1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}$ .

Equivalently, we have

$1 > \frac{\sin \theta}{\theta} > \cos \theta$ .

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

$\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}=1$ .

In (Effigy) we use this limit to establish $\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0$ . This limit too proves useful in later capacity.

Evaluating an Important Trigonometric Limit

Evaluate $\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}$ .

Solution

In the starting time pace, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in the numerator to a sine:

$\begin{array}{cc} \underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}& =\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta} \cdot \frac{1+ \cos \theta}{1+ \cos \theta} \\ & =\underset{\theta \to 0}{\lim}\frac{1-\cos^2 \theta}{\theta(1+ \cos \theta)} \\ & =\underset{\theta \to 0}{\lim}\frac{\sin^2 \theta}{\theta(1+ \cos \theta)} \\ & =\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{1+ \cos \theta} \\ & =1 \cdot \frac{0}{2}=0 \end{array}$

Therefore,

$\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0$ .

Evaluate $\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\sin \theta}$ .

Key Concepts

The limit laws permit us to evaluate limits of functions without having to go through step-by-pace processes each time.
For polynomials and rational functions, $\underset{x\to a}{\lim}f(x)=f(a)$ .
Yous can evaluate the limit of a role by factoring and canceling, by multiplying by a conjugate, or past simplifying a complex fraction.
The Squeeze Theorem allows you lot to detect the limit of a function if the part is always greater than one function and less than another function with limits that are known.

Primal Equations

In the post-obit exercises, use the limit laws to evaluate each limit. Justify each step by indicating the advisable limit law(s).

i. $\underset{x\to 0}{\lim}(4x^2-2x+3)$

Solution

Utilise abiding multiple police and difference law: $\underset{x\to 0}{\lim}(4x^2-2x+3)=4\underset{x\to 0}{\lim}x^2-2\underset{x\to 0}{\lim}x+\underset{x\to 0}{\lim}3=3$

two. $\underset{x\to 1}{\lim}\frac{x^3+3x^2+5}{4-7x}$

iii. $\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}$

Solution

Use root law: $\underset{x\to -2}{\lim}\sqrt{x^2-6x+3}=\sqrt{\underset{x\to -2}{\lim}(x^2-6x+3)}=\sqrt{19}$

4. $\underset{x\to -1}{\lim}(9x+1)^2$

In the post-obit exercises, utilise direct substitution to evaluate each limit.

5. $\underset{x\to 7}{\lim}x^2$

six. $\underset{x\to -2}{\lim}(4x^2-1)$

seven. $\underset{x\to 0}{\lim}\frac{1}{1+ \sin x}$

eight. $\underset{x\to 2}{\lim}e^{2x-x^2}$

nine. $\underset{x\to 1}{\lim}\frac{2-7x}{x+6}$

Solution

$-\frac{5}{7}$

10. $\underset{x\to 3}{\lim}\ln e^{3x}$

In the following exercises, use direct substitution to show that each limit leads to the indeterminate form 0/0. Then, evaluate the limit.

12. $\underset{x\to 2}{\lim}\frac{x-2}{x^2-2x}$

14. $\underset{h\to 0}{\lim}\frac{(1+h)^2-1}{h}$

xvi. $\underset{h\to 0}{\lim}\frac{\frac{1}{a+h}-\frac{1}{a}}{h}$ , where $a$ is a existent-valued abiding

xviii. $\underset{x\to 1}{\lim}\frac{x^3-1}{x^2-1}$

20. $\underset{x\to -3}{\lim}\frac{\sqrt{x+4}-1}{x+3}$

In the following exercises, apply direct substitution to obtain an undefined expression. Then, use the method of (Figure) to simplify the role to aid determine the limit.

21. $\underset{x\to -2^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

Solution

$-\infty$

22. $\underset{x\to -2^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

23. $\underset{x\to 1^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

Solution

$-\infty$

24. $\underset{x\to 1^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}$

In the following exercises, assume that $\underset{x\to 6}{\lim}f(x)=4, \, \underset{x\to 6}{\lim}g(x)=9$ , and $\underset{x\to 6}{\lim}h(x)=6$ . Use these three facts and the limit laws to evaluate each limit.

25. $\underset{x\to 6}{\lim}2f(x)g(x)$

Solution

$\underset{x\to 6}{\lim}2f(x)g(x)=2\underset{x\to 6}{\lim}f(x)\underset{x\to 6}{\lim}g(x)=72$

26. $\underset{x\to 6}{\lim}\frac{g(x)-1}{f(x)}$

27. $\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))$

Solution

$\underset{x\to 6}{\lim}(f(x)+\frac{1}{3}g(x))=\underset{x\to 6}{\lim}f(x)+\frac{1}{3}\underset{x\to 6}{\lim}g(x)=7$

28. $\underset{x\to 6}{\lim}\frac{(h(x))^3}{2}$

29. $\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}$

Solution

$\underset{x\to 6}{\lim}\sqrt{g(x)-f(x)}=\sqrt{\underset{x\to 6}{\lim}g(x)-\underset{x\to 6}{\lim}f(x)}=\sqrt{5}$

xxx. $\underset{x\to 6}{\lim}x \cdot h(x)$

31. $\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]$

Solution

$\underset{x\to 6}{\lim}[(x+1)\cdot f(x)]=(\underset{x\to 6}{\lim}(x+1))(\underset{x\to 6}{\lim}f(x))=28$ .

32. $\underset{x\to 6}{\lim}(f(x) \cdot g(x)-h(x))$

In the following exercises, utilize a calculator to describe the graph of each piecewise-divers function and study the graph to evaluate the given limits.

Solution

iii. There is an open circle at (3, seven), and the slope is ane.">
a. 9; b. vii

Solution

= 2. It has a slope of -1 and an x intercept at (three,0).">
a. one; b. 1

In the following exercises, employ the following graphs and the limit laws to evaluate each limit.

-3. Other central points are (0, 1), (-v,2), (i,2), (-7, 4), and (-nine,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing gradient. Information technology goes to (-3,-two) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a piddling below y=-2. Other key points are (0, -vii/three), (-five,0), (1,-5), (-7, 2), and (-9, 4).">

36. $\underset{x\to -3^+}{\lim}(f(x)+g(x))$

37. $\underset{x\to -3^-}{\lim}(f(x)-3g(x))$

Solution

$\underset{x\to -3^-}{\lim}(f(x)-3g(x))=\underset{x\to -3^-}{\lim}f(x)-3\underset{x\to -3^-}{\lim}g(x)=0+6=6$

38. $\underset{x\to 0}{\lim}\frac{f(x)g(x)}{3}$

39. $\underset{x\to -5}{\lim}\frac{2+g(x)}{f(x)}$

Solution

$\underset{x\to -5}{\lim}\frac{2+g(x)}{f(x)}=\frac{2+(\underset{x\to -5}{\lim}g(x))}{\underset{x\to -5}{\lim}f(x)}=\frac{2+0}{2}=1$

40. $\underset{x\to 1}{\lim}(f(x))^2$

41. $\underset{x\to 1}{\lim}\sqrt{f(x)-g(x)}$

[reveal-respond q="957757″]Show Solution[/reveal-answer]
[hidden-answer a="957757″]
$\underset{x\to 1}{\lim}\sqrt[3]{f(x)-g(x)}=\sqrt[3]{\underset{x\to 1}{\lim}f(x)-\underset{x\to 1}{\lim}g(x)}=\sqrt[3]{2+5}=\sqrt[3]{7}$

42. $\underset{x\to -7}{\lim}(x \cdot g(x))$

43. $\underset{x\to -9}{\lim}[xf(x)+2g(x)]$

Solution

$\underset{x\to -9}{\lim}(xf(x)+2g(x))=(\underset{x\to -9}{\lim}x)(\underset{x\to -9}{\lim}f(x))+2\underset{x\to -9}{\lim}(g(x))=(-9)(6)+2(4)=-46$

For the following problems, evaluate the limit using the Clasp Theorem. Use a calculator to graph the functions $f(x), \, g(x)$ , and $h(x)$ when possible.

44. [T] True or Simulated? If $2x-1\le g(x)\le x^2-2x+3$ , then $\underset{x\to 2}{\lim}g(x)=0$ .

45. [T] $\underset{\theta \to 0}{\lim}\theta^2 \cos(\frac{1}{\theta})$

Solution

The limit is zero.

The graph of three functions over the domain [-1,1], colored red, green, and blue as follows: red: theta^2, green: theta^2 * cos (1/theta), and blue: - (theta^2). The red and blue functions open upwards and downwards respectively as parabolas with vertices at the origin. The green function is trapped between the two.

46. $\underset{x\to 0}{\lim}f(x)$ , where $f(x)=\begin{cases} 0 & x \, \text{rational} \\ x^2 & x \, \text{irrational} \end{cases}$

Solution

A graph of a function with two curves. The first is in quadrant two and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to negative infinity. The second is in quadrant one and curves asymptotically to infinity along the y axis and to 0 along the x axis as x goes to infinity.
b. $\underset{r\to 0^+}{\lim}E(r)=\infty$ . The magnitude of the electric field as yous arroyo the particle $q$ becomes space. It does non make physical sense to evaluate negative distance.

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Source: https://opentextbc.ca/calculusv1openstax/chapter/the-limit-laws/

If F(X) Is a Polynomial Then Lim F(X)=f(C) True/false Homework Exercises Review

2.3 The Limit Laws

Learning Objectives

Evaluating Limits with the Limit Laws

Basic Limit Results

Evaluating a Basic Limit

Evaluating a Limit Using Limit Laws

Solution

Using Limit Laws Repeatedly

Solution

Solution

Limits of Polynomial and Rational Functions

Limits of Polynomial and Rational Functions

Evaluating a Limit of a Rational Function

Solution

Additional Limit Evaluation Techniques

Evaluating a Limit by Factoring and Canceling

Solution

Solution

Evaluating a Limit by Multiplying by a Conjugate

Solution

Solution

Evaluating a Limit by Simplifying a Complex Fraction

Solution

Evaluating a Limit When the Limit Laws Do Non Apply

Solution

Solution

Evaluating a 1-Sided Limit Using the Limit Laws

Evaluating a Two-Sided Limit Using the Limit Laws

Solution

Evaluating a Limit of the Form Using the Limit Laws

Solution

The Squeeze Theorem

The Squeeze Theorem

Applying the Squeeze Theorem

Evaluating an Important Trigonometric Limit

Solution

Key Concepts

Primal Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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Evaluating a Limit of the Form $K/0, \, K\ne 0$ Using the Limit Laws