The Limit Laws · Calculus

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 learn how to apply these laws. In the Student Project at the end of this section, you have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two results, together with the limit laws, serve as a foundation for calculating many limits.

Evaluating Limits with the Limit Laws

The first two limit laws were stated in [link] 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.

Limit Laws

Let $f (x)$

and $g (x)$

be defined for all $x \neq a$

over some open interval containing a. Assume that L and M are real numbers such that $lim_{x \to a} f (x) = L$

and $lim_{x \to a} g (x) = M .$

Let c be a constant. Then, each of the following statements holds:

Sum law for limits: $lim_{x \to a} (f (x) + g (x)) = lim_{x \to a} f (x) + lim_{x \to a} g (x) = L + M$

Difference law for limits: $lim_{x \to a} (f (x) - g (x)) = lim_{x \to a} f (x) - lim_{x \to a} g (x) = L - M$

Constant multiple law for limits: $lim_{x \to a} c f (x) = c \cdot lim_{x \to a} f (x) = c L$

Product law for limits: $lim_{x \to a} (f (x) \cdot g (x)) = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x) = L \cdot M$

Quotient law for limits: $lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)} = \frac{L}{M}$

for $M \neq 0$

Power law for limits: $lim_{x \to a} {(f (x))}^{n} = {(lim_{x \to a} f (x))}^{n} = L^{n}$

for every positive integer n.

Root law for limits: $lim_{x \to a} \sqrt[n]{f (x)} = \sqrt[n]{lim_{x \to a} f (x)} = \sqrt[n]{L}$

for all L if n is odd and for $L \geq 0$

if n is even.

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

lim_{x \to a} f (x) = f (a) .

This is not always true, but it 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.

To see that this theorem holds, consider the polynomial

p (x) = c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0} .

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

lim_{x \to a} f (x)

is undefined. The following observation allows us to evaluate many limits of this type:

over some open interval containing a, then

lim_{x \to a} f (x) = lim_{x \to a} g (x) .

To understand this idea better, consider the limit

lim_{x \to 1} \frac{x^{2} - 1}{x - 1} .

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. [link] illustrates the factor-and-cancel technique; [link] shows multiplying by a conjugate. In [link], we look at simplifying a complex fraction.

[link] 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

lim_{x \to a^{-}} h (x),

In [link] 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.

We now turn our attention to evaluating a limit of the form

lim_{x \to a} \frac{f (x)}{g (x)},

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 a that is unknown, between two functions having a common known limit at a. [link] illustrates this idea.

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

lim_{θ \to 0} sin θ .

is the y-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

0 < θ < \frac{π}{2}, 0 < sin θ < θ .

An application of the squeeze theorem produces the desired limit. Thus, since

lim_{θ \to 0^{+}} sin θ = 0

We now take a look at a limit that plays an important role in later chapters—namely,

lim_{θ \to 0} \frac{sin θ}{θ} .

To evaluate this limit, we use the unit circle in [link]. Notice that this figure adds one additional triangle to [link]. We see that the length of the side opposite angle θ in this new triangle is

tan θ .

By applying a manipulation similar to that used in demonstrating that

lim_{θ \to 0^{-}} sin θ = 0,

In [link] we use this limit to establish

lim_{θ \to 0} \frac{1 - cos θ}{θ} = 0 .

Deriving the Formula for the Area of a Circle

Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased. He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have predicted about the limit.

We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular polygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you can obtain the area of the circle. To see this, carry out the following steps:

Express the height h and the base b of the isosceles triangle in [link] in terms of $θ$
and r.
Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r.

(Substitute
$(1 / 2) sin θ$
for
$sin (θ / 2) cos (θ / 2)$
in your expression.)
If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve this for n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)
Find an expression for the area of the n-sided polygon in terms of r and θ.
To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. (Hint: $lim_{θ \to 0} \frac{(sin θ)}{θ} = 1).$

The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration.

Key Concepts

Key Equations

In the following exercises, use the limit laws to evaluate each limit. Justify each step by indicating the appropriate limit law(s).

lim_{x \to 0} (4 x^{2} - 2 x + 3)

Use constant multiple law and difference law: $lim_{x \to 0} (4 x^{2} - 2 x + 3) = 4 lim_{x \to 0} x^{2} - 2 lim_{x \to 0} x + lim_{x \to 0} 3 = 3$

lim_{x \to 1} \frac{x^{3} + 3 x^{2} + 5}{4 - 7 x}

lim_{x \to −2} \sqrt{x^{2} - 6 x + 3}

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

lim_{x \to −1} {(9 x + 1)}^{2}

In the following exercises, use direct substitution to evaluate each limit.

lim_{x \to 7} x^{2}

lim_{x \to −2} (4 x^{2} - 1)

lim_{x \to 0} \frac{1}{1 + sin x}

lim_{x \to 2} e^{2 x - x^{2}}

lim_{x \to 1} \frac{2 - 7 x}{x + 6}

- \frac{5}{7}

lim_{x \to 3} ln e^{3 x}

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

Then, evaluate the limit.

lim_{x \to 4} \frac{x^{2} - 16}{x - 4}

lim_{x \to 4} \frac{x^{2} - 16}{x - 4} = \frac{16 - 16}{4 - 4} = \frac{0}{0};

then, $lim_{x \to 4} \frac{x^{2} - 16}{x - 4} = lim_{x \to 4} \frac{(x + 4) (x - 4)}{x - 4} = 8$

lim_{x \to 2} \frac{x - 2}{x^{2} - 2 x}

lim_{x \to 6} \frac{3 x - 18}{2 x - 12}

lim_{x \to 6} \frac{3 x - 18}{2 x - 12} = \frac{18 - 18}{12 - 12} = \frac{0}{0};

then, $lim_{x \to 6} \frac{3 x - 18}{2 x - 12} = lim_{x \to 6} \frac{3 (x - 6)}{2 (x - 6)} = \frac{3}{2}$

lim_{h \to 0} \frac{{(1 + h)}^{2} - 1}{h}

lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3}

lim_{x \to 9} \frac{t - 9}{\sqrt{t} - 3} = \frac{9 - 9}{3 - 3} = \frac{0}{0};

then, $lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3} = lim_{t \to 9} \frac{t - 9}{\sqrt{t} - 3} \frac{\sqrt{t} + 3}{\sqrt{t} + 3} = lim_{t \to 9} (\sqrt{t} + 3) = 6$

lim_{h \to 0} \frac{\frac{1}{a + h} - \frac{1}{a}}{h},

where a is a real-valued constant

lim_{θ \to π} \frac{sin θ}{tan θ}

lim_{θ \to π} \frac{sin θ}{tan θ} = \frac{sin π}{tan π} = \frac{0}{0};

then, $lim_{θ \to π} \frac{sin θ}{tan θ} = lim_{θ \to π} \frac{sin θ}{\frac{sin θ}{cos θ}} = lim_{θ \to π} cos θ = −1$

lim_{x \to 1} \frac{x^{3} - 1}{x^{2} - 1}

lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1}

lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1} = \frac{\frac{1}{2} + \frac{3}{2} - 2}{1 - 1} = \frac{0}{0};

then, $lim_{x \to 1 / 2} \frac{2 x^{2} + 3 x - 2}{2 x - 1} = lim_{x \to 1 / 2} \frac{(2 x - 1) (x + 2)}{2 x - 1} = \frac{5}{2}$

lim_{x \to −3} \frac{\sqrt{x + 4} - 1}{x + 3}

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of [link] to simplify the function to help determine the limit.

lim_{x \to {−2}^{-}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}

−∞

lim_{x \to {−2}^{+}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}

lim_{x \to 1^{-}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}

−∞

lim_{x \to 1^{+}} \frac{2 x^{2} + 7 x - 4}{x^{2} + x - 2}

In the following exercises, assume that $lim_{x \to 6} f (x) = 4, lim_{x \to 6} g (x) = 9,$

and $lim_{x \to 6} h (x) = 6 .$

Use these three facts and the limit laws to evaluate each limit.

lim_{x \to 6} 2 f (x) g (x)

lim_{x \to 6} 2 f (x) g (x) = 2 lim_{x \to 6} f (x) lim_{x \to 6} g (x) = 72

lim_{x \to 6} \frac{g (x) - 1}{f (x)}

lim_{x \to 6} (f (x) + \frac{1}{3} g (x))

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

lim_{x \to 6} \frac{{(h (x))}^{3}}{2}

lim_{x \to 6} \sqrt{g (x) - f (x)}

lim_{x \to 6} \sqrt{g (x) - f (x)} = \sqrt{lim_{x \to 6} g (x) - lim_{x \to 6} f (x)} = \sqrt{5}

lim_{x \to 6} x \cdot h (x)

lim_{x \to 6} [(x + 1) \cdot f (x)]

lim_{x \to 6} [(x + 1) f (x)] = (lim_{x \to 6} (x + 1)) (lim_{x \to 6} f (x)) = 28

lim_{x \to 6} (f (x) \cdot g (x) - h (x))

[T] In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

f (x) = {\begin{cases} x^{2}, & x \leq 3 \\ x + 4, & x > 3 \end{cases}

$lim_{x \to 3^{-}} f (x)$
$lim_{x \to 3^{+}} f (x)$

The graph of a piecewise function with two segments. The first is the parabola x^2, which exists for x<=3. The vertex is at the origin, it opens upward, and there is a closed circle at the endpoint (3,9). The second segment is the line x+4, which is a linear function existing for x > 3. There is an open circle at (3, 7), and the slope is 1.

a. 9; b. 7

g (x) = {\begin{cases} x^{3} - 1, & x \leq 0 \\ 1, & x > 0 \end{cases}

$lim_{x \to 0^{-}} g (x)$
$lim_{x \to 0^{+}} g (x)$

h (x) = {\begin{cases} x^{2} - 2 x + 1, & x < 2 \\ 3 - x, & x \geq 2 \end{cases}

$lim_{x \to 2^{-}} h (x)$
$lim_{x \to 2^{+}} h (x)$

The graph of a piecewise function with two segments. The first segment is the parabola x^2 – 2x + 1, for x < 2. It opens upward and has a vertex at (1,0). The second segment is the line 3-x for x>= 2. It has a slope of -1 and an x intercept at (3,0).

a. 1; b. 1

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

lim_{x \to {−3}^{+}} (f (x) + g (x))

lim_{x \to {−3}^{-}} (f (x) - 3 g (x))

lim_{x \to {−3}^{-}} (f (x) - 3 g (x)) = lim_{x \to {−3}^{-}} f (x) - 3 lim_{x \to {−3}^{-}} g (x) = 0 + 6 = 6

lim_{x \to 0} \frac{f (x) g (x)}{3}

lim_{x \to −5} \frac{2 + g (x)}{f (x)}

lim_{x \to −5} \frac{2 + g (x)}{f (x)} = \frac{2 + (lim_{x \to −5} g (x))}{lim_{x \to −5} f (x)} = \frac{2 + 0}{2} = 1

lim_{x \to 1} {(f (x))}^{2}

lim_{x \to 1} \sqrt{f (x) - g (x)}

lim_{x \to 1} \sqrt[3]{f (x) - g (x)} = \sqrt[3]{lim_{x \to 1} f (x) - lim_{x \to 1} g (x)} = \sqrt[3]{2 + 5} = \sqrt[3]{7}

lim_{x \to −7} (x \cdot g (x))

lim_{x \to −9} [x \cdot f (x) + 2 \cdot g (x)]

lim_{x \to −9} (x f (x) + 2 g (x)) = (lim_{x \to −9} x) (lim_{x \to −9} f (x)) + 2 lim_{x \to −9} (g (x)) = (−9) (6) + 2 (4) = −46

For the following problems, evaluate the limit using the squeeze theorem. Use a calculator to graph the functions $f (x), g (x),$

and $h (x)$

when possible.

[T] True or False? If $2 x - 1 \leq g (x) \leq x^{2} - 2 x + 3,$

then $lim_{x \to 2} g (x) = 0 .$

[T] $lim_{θ \to 0} θ^{2} cos (\frac{1}{θ})$

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.

lim_{x \to 0} f (x),

where $f (x) = {\begin{cases} 0, & x rational \\ x^{2}, & x irrrational \end{cases}$

[T] In physics, the magnitude of an electric field generated by a point charge at a distance r in vacuum is governed by Coulomb’s law: $E (r) = \frac{q}{4 π ε_{0} r^{2}},$

where E represents the magnitude of the electric field, q is the charge of the particle, r is the distance between the particle and where the strength of the field is measured, and $\frac{1}{4 π ε_{0}}$

is Coulomb’s constant: $8.988 \times 10^{9} N \cdot m^{2} / C^{2} .$

Use a graphing calculator to graph $E (r)$
given that the charge of the particle is
$q = 10^{−10} .$
Evaluate $lim_{r \to 0^{+}} E (r) .$
What is the physical meaning of this quantity? Is it physically relevant? Why are you evaluating from the right?

a.* * *

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. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

[T] The density of an object is given by its mass divided by its volume: $ρ = m / V .$

Use a calculator to plot the volume as a function of density $(V = m / ρ),$
assuming you are examining something of mass 8 kg (
$m = 8).$
Evaluate $lim_{ρ \to 0^{+}} V (ρ)$
and explain the physical meaning.

Glossary

constant multiple law for limits: the limit law $lim_{x \to a} c f (x) = c \cdot lim_{x \to a} f (x) = c L$

difference law for limits: the limit law $lim_{x \to a} (f (x) - g (x)) = lim_{x \to a} f (x) - lim_{x \to a} g (x) = L - M$

limit laws

the individual properties of limits; for each of the individual laws, let

f (x)

and

g (x)

be defined for all

x \neq a

over some open interval containing a; assume that L and M are real numbers so that

lim_{x \to a} f (x) = L

and

lim_{x \to a} g (x) = M;

let c be a constant

power law for limits: the limit law $lim_{x \to a} {(f (x))}^{n} = {(lim_{x \to a} f (x))}^{n} = L^{n}$
for every positive integer n

product law for limits: the limit law $lim_{x \to a} (f (x) \cdot g (x)) = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x) = L \cdot M$

quotient law for limits: the limit law $lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)} = \frac{L}{M}$
for
$M \neq 0$

root law for limits

the limit law

lim_{x \to a} \sqrt[n]{f (x)} = \sqrt[n]{lim_{x \to a} f (x)} = \sqrt[n]{L}

for all L if n is odd and for

L \geq 0

if n is even

squeeze theorem

states that if

f (x) \leq g (x) \leq h (x)

for all

x \neq a

over an open interval containing a and

lim_{x \to a} f (x) = L = lim_{x \to a} h (x)

where L is a real number, then

lim_{x \to a} g (x) = L

sum law for limits: The limit law $lim_{x \to a} (f (x) + g (x)) = lim_{x \to a} f (x) + lim_{x \to a} g (x) = L + M$