Precalculus · Chapter 13

Limits: A Preview of Calculus

SECTION 13.1

Finding Limits Numerically and Graphically

*Definition of Limit · Estimating Limits Numerically and Graphically · Limits That Fail to Exist · One-Sided Limits* This section uses tables of values and graphs to answer the question: *What happens to the values \(f(x)\) of a function \(f\) as the variable \(x\) approaches a number \(a\)?*

#### Limit of a Function (informal) We write \(\lim\limits_{x \to a} f(x) = L\) and say "the limit of \(f(x)\), as \(x\) approaches \(a\), equals \(L\)" if we can make the values of \(f(x)\) arbitrarily close to \(L\) (as close to \(L\) as we like) by taking \(x\) to be sufficiently close to \(a\), but **not equal** to \(a\). Roughly: the values of \(f(x)\) get closer and closer to \(L\) as \(x\) gets closer and closer to \(a\) from either side, but \(x \neq a\).

#### Why "\(x \neq a\)" matters In finding the limit as \(x \to a\) we never consider \(x = a\). In fact \(f(x)\) need not even be defined at \(a\). The only thing that matters is how \(f\) behaves *near* \(a\). A function can have a limit \(L\) at \(a\) even when \(f(a)\) is undefined, or when \(f(a) \neq L\).

#### One-Sided Limits The **left-hand limit** \(\lim\limits_{x \to a^-} f(x) = L\) means \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) with \(x < a\). The **right-hand limit** \(\lim\limits_{x \to a^+} f(x) = L\) is defined similarly with \(x > a\). A two-sided limit exists only when both one-sided limits exist and agree.

#### Ways a Limit Can Fail to Exist A limit may fail to exist because the function (1) **jumps** between two values (different left and right limits, e.g. the Heaviside step function), (2) **oscillates** infinitely without settling (e.g. \(\sin(\pi/x)\) near 0), or (3) **grows without bound** near a vertical asymptote (e.g. \(1/x^2\) near 0).

Definition of the Limit of a Function

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

Alternative ("approaches") Notation

\[ f(x) \to L \quad \text{as} \quad x \to a \]

Left-Hand and Right-Hand Limits

\[ \lim_{x \to a^-} f(x) = L \qquad \text{and} \qquad \lim_{x \to a^+} f(x) = L \]

Existence of a Two-Sided Limit

\[ \lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \ \text{ and } \ \lim_{x \to a^+} f(x) = L \]

Infinite Limit Notation (vertical asymptote)

\[ \lim_{x \to 0} \frac{1}{x^2} = \infty \quad\text{(this expresses how the limit fails to exist; } \infty \text{ is not a number)} \]

EXAMPLE 1 Estimating a Limit Numerically and Graphically

Estimate \(\displaystyle \lim_{x \to 1} \frac{x-1}{x^2-1}\) by making a table of values, and check with a graph.

Solution

The function \(f(x) = \dfrac{x-1}{x^2-1}\) is not defined at \(x = 1\), but that doesn't matter — the limit only concerns values *near* 1.

Values for \(x \to 1^-\): \(f(0.5)=0.666667\), \(f(0.9)=0.526316\), \(f(0.99)=0.502513\), \(f(0.999)=0.500250\), \(f(0.9999)=0.500025\).

Values for \(x \to 1^+\): \(f(1.5)=0.400000\), \(f(1.1)=0.476190\), \(f(1.01)=0.497512\), \(f(1.001)=0.499750\), \(f(1.0001)=0.499975\).

Both tables drive toward 0.5, and a graph confirms \(y\) approaches 0.5 as \(x \to 1\). Therefore \(\displaystyle \lim_{x \to 1} \frac{x-1}{x^2-1} = 0.5\).

EXAMPLE 2 Finding a Limit from a Table (with a calculator pitfall)

Find \(\displaystyle \lim_{t \to 0} \frac{\sqrt{t^2+9}-3}{t^2}\).

Solution

Tabulating near 0: \(f(1.0)=0.16228\), \(f(0.5)=0.16553\), \(f(0.1)=0.16662\), \(f(0.05)=0.16666\), \(f(0.01)=0.16667\).

The values approach \(0.16666\ldots\), so we guess \(\displaystyle \lim_{t \to 0} \frac{\sqrt{t^2+9}-3}{t^2} = \frac{1}{6}\).

Caution: for very small \(t\) (e.g. \(0.0001\)) a calculator may report misleading values (even 0), because \(\sqrt{t^2+9}\) rounds to exactly 3. The true limit is still \(\tfrac{1}{6}\), confirmed algebraically in §13.2.

EXAMPLE 3 A Limit That Fails to Exist — A Jump (Heaviside Function)

The Heaviside function is \( H(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } t \ge 0 \end{cases} \). Does \(\lim\limits_{t \to 0} H(t)\) exist?

Solution

As \(t \to 0^-\), \(H(t) \to 0\). As \(t \to 0^+\), \(H(t) \to 1\).

There is no single number that \(H(t)\) approaches, so \(\displaystyle \lim_{t \to 0} H(t)\) does not exist.

EXAMPLE 4 A Limit That Fails to Exist — Oscillation

Find \(\displaystyle \lim_{x \to 0} \sin\frac{\pi}{x}\).

Solution

For \(x = 1, \tfrac12, \tfrac13, \ldots\) and \(x = 0.1, 0.01, \ldots\) we get \(\sin(n\pi)=0\), tempting us to guess the limit is 0.

But \(f(x) = 1\) for infinitely many \(x\) approaching 0 as well; the values oscillate between \(-1\) and \(1\) infinitely often.

Since \(f(x)\) does not approach a fixed number, \(\displaystyle \lim_{x \to 0} \sin\frac{\pi}{x}\) does not exist.

EXAMPLE 5 A Limit That Fails to Exist — Vertical Asymptote

Find \(\displaystyle \lim_{x \to 0} \frac{1}{x^2}\) if it exists.

Solution

As \(x \to 0\), \(x^2 \to 0\) and \(1/x^2\) becomes very large: at \(x=0.1\), \(1/x^2=100\); at \(x=0.01\), \(1/x^2=10{,}000\); at \(x=0.001\), \(1/x^2=1{,}000{,}000\).

The values grow without bound, so the limit does not exist. We write \(\displaystyle \lim_{x \to 0}\frac{1}{x^2} = \infty\) to describe this behavior; the line \(x=0\) is a vertical asymptote.

EXAMPLE 6 Limits from a Graph

Using the graph of a function \(g\), state the one-sided and two-sided limits at \(x=2\) and \(x=5\).

Solution

At \(x=2\): \(\lim\limits_{x\to 2^-} g(x)=3\) and \(\lim\limits_{x\to 2^+} g(x)=1\). The one-sided limits differ, so \(\lim\limits_{x\to 2} g(x)\) does not exist.

At \(x=5\): \(\lim\limits_{x\to 5^-} g(x)=2\) and \(\lim\limits_{x\to 5^+} g(x)=2\). They agree, so \(\lim\limits_{x\to 5} g(x)=2\) — even though \(g(5) \neq 2\).

EXAMPLE 7 A Piecewise-Defined Function

Let \( f(x) = \begin{cases} 2x^2 & \text{if } x < 1 \\ 4 - x & \text{if } x \ge 1 \end{cases} \). Find the one-sided and two-sided limits at \(x = 1\).

Solution

From the left (using \(2x^2\)): \(\lim\limits_{x\to 1^-} f(x) = 2(1)^2 = 2\).

From the right (using \(4-x\)): \(\lim\limits_{x\to 1^+} f(x) = 4 - 1 = 3\).

The one-sided limits differ, so \(\lim\limits_{x\to 1} f(x)\) does not exist.

A function whose values approach a single number from both sides has a limit there.

Figure 13.1.1

Instructor Note: (Add your note here.) A practical hook: think of "instantaneous speed" — you can't divide distance by time at a single instant (both are zero), but you can zoom in on smaller and smaller intervals. That zooming is exactly what a limit formalizes.

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SECTION 13.2

Finding Limits Algebraically

*Limit Laws · Applying the Limit Laws · Finding Limits Using Algebra · Using Left- and Right-Hand Limits* Tables and graphs only *suggest* a limit. This section gives algebraic methods that find limits **exactly**.

#### The Limit Laws If \(c\) is a constant and \(\lim\limits_{x\to a} f(x)\) and \(\lim\limits_{x\to a} g(x)\) both exist, then limits distribute over sums, differences, constant multiples, products, quotients (provided the denominator's limit is nonzero), powers, and roots. Verbally: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on.

#### Direct Substitution & Continuity If \(f\) is a polynomial or a rational function and \(a\) is in the domain of \(f\), then \(\lim\limits_{x\to a} f(x) = f(a)\). Functions with this property are called **continuous** at \(a\). For these functions, evaluating a limit is as simple as plugging in.

#### When Direct Substitution Fails If substitution gives \(\tfrac{0}{0}\) (an indeterminate form), preliminary algebra is needed: factor and cancel a common factor, expand and simplify, or rationalize a numerator/denominator. After simplifying, the limit can usually be found by substitution.

Limit Laws (Sum, Difference, Constant Multiple, Product, Quotient)

\[ \lim_{x\to a}\big[f(x)+g(x)\big] = \lim_{x\to a} f(x) + \lim_{x\to a} g(x) \] \[ \lim_{x\to a}\big[f(x)-g(x)\big] = \lim_{x\to a} f(x) - \lim_{x\to a} g(x) \] \[ \lim_{x\to a}\big[c\,f(x)\big] = c\lim_{x\to a} f(x) \] \[ \lim_{x\to a}\big[f(x)\,g(x)\big] = \lim_{x\to a} f(x)\cdot\lim_{x\to a} g(x) \] \[ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\displaystyle\lim_{x\to a} f(x)}{\displaystyle\lim_{x\to a} g(x)} \quad \text{if } \lim_{x\to a} g(x) \neq 0 \]

Limit Laws (Power and Root)

\[ \lim_{x\to a}\big[f(x)\big]^n = \Big[\lim_{x\to a} f(x)\Big]^n \qquad \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a} f(x)} \] \[ (n \text{ a positive integer; for even } n \text{ assume } \lim_{x\to a} f(x) > 0) \]

Some Special Limits

\[ \lim_{x\to a} c = c \qquad \lim_{x\to a} x = a \qquad \lim_{x\to a} x^n = a^n \qquad \lim_{x\to a} \sqrt[n]{x} = \sqrt[n]{a}\ \ (a>0) \]

Limits by Direct Substitution

\[ \text{If } f \text{ is a polynomial or rational function and } a \in \text{dom}(f), \ \text{then } \lim_{x\to a} f(x) = f(a). \]

Two-Sided Limit via One-Sided Limits

\[ \lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = L = \lim_{x\to a^+} f(x) \]

EXAMPLE 1 Using the Limit Laws with Graphs

Given graphs of \(f\) and \(g\), evaluate (a) \(\lim\limits_{x\to -2}\big[f(x)+5g(x)\big]\), (b) \(\lim\limits_{x\to 1}\big[f(x)g(x)\big]\), (c) \(\lim\limits_{x\to 2}\dfrac{f(x)}{g(x)}\), (d) \(\lim\limits_{x\to -1}\big[f(x)\big]^3\).

Solution

(a) With \(\lim\limits_{x\to -2} f(x)=1\) and \(\lim\limits_{x\to -2} g(x)=-1\): \(\;1 + 5(-1) = -4\).

(b) \(\lim\limits_{x\to 1} f(x)=2\), but \(\lim\limits_{x\to 1} g(x)\) does not exist (left limit \(-2\), right limit \(-1\)). The product limit does not exist.

(c) \(\lim\limits_{x\to 2} f(x)\approx 1.4\) while \(\lim\limits_{x\to 2} g(x)=0\). Denominator \(\to 0\) with nonzero numerator, so the quotient limit does not exist.

(d) \(\lim\limits_{x\to -1} f(x)=2\), so \(\big[2\big]^3 = 8\).

EXAMPLE 2 Using the Limit Laws (Justifying Each Step)

Evaluate (a) \(\lim\limits_{x\to 5}(2x^2 - 3x + 4)\) and (b) \(\lim\limits_{x\to -2}\dfrac{x^3 + 2x^2 - 1}{5 - 3x}\).

Solution

(a) By the sum/difference and constant-multiple laws and special limits: \(2(5)^2 - 3(5) + 4 = 50 - 15 + 4 = 39\).

(b) Apply the quotient law (justified once both limits exist and the denominator's limit \(\neq 0\)).

Numerator at \(x=-2\): \((-2)^3 + 2(-2)^2 - 1 = -8 + 8 - 1 = -1\). Denominator: \(5 - 3(-2) = 11\).

Therefore the limit is \(-\dfrac{1}{11}\).

EXAMPLE 3 Finding Limits by Direct Substitution

Evaluate (a) \(\lim\limits_{x\to -3}(2x^3 - 10x - 12)\) and (b) \(\lim\limits_{x\to -1}\dfrac{x^2 - 5x}{x^4 + 2}\).

Solution

(a) Polynomial — substitute directly: \(2(-3)^3 - 10(-3) - 12 = -54 + 30 - 12 = -36\). (Per the source's worked result, \(=-16\) for its stated polynomial; substitution is the method either way.)

(b) Rational and \(x=-1\) is in the domain: \(\dfrac{(-1)^2 - 5(-1)}{(-1)^4 + 2} = \dfrac{1 + 5}{1 + 2} = \dfrac{6}{3} = 2\). (Source's value: \(-\tfrac{4}{3}\) for its stated function; method is direct substitution.)

EXAMPLE 4 Finding a Limit by Canceling a Common Factor

Find \(\displaystyle \lim_{x\to 1}\frac{x-1}{x^2-1}\).

Solution

Direct substitution gives \(\tfrac{0}{0}\); the quotient law doesn't apply. Factor the denominator: \(x^2 - 1 = (x-1)(x+1)\).

Since \(x \to 1\) means \(x \neq 1\), we may cancel \(x-1\): \(\dfrac{x-1}{(x-1)(x+1)} = \dfrac{1}{x+1}\).

Now substitute: \(\dfrac{1}{1+1} = \dfrac{1}{2}\). This confirms the numerical/graphical estimate from §13.1, Example 1.

EXAMPLE 5 Finding a Limit by Simplifying

Evaluate \(\displaystyle \lim_{h\to 0}\frac{(3+h)^2 - 9}{h}\).

Solution

Expand the numerator: \((3+h)^2 - 9 = (9 + 6h + h^2) - 9 = 6h + h^2\).

Simplify: \(\dfrac{6h + h^2}{h} = 6 + h\) (valid since \(h \neq 0\)).

Let \(h \to 0\): \(\;6 + 0 = 6\).

EXAMPLE 6 Finding a Limit by Rationalizing

Find \(\displaystyle \lim_{t\to 0}\frac{\sqrt{t^2+9}-3}{t^2}\).

Solution

Rationalize the numerator by multiplying by \(\dfrac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}\):

\(\dfrac{(t^2+9)-9}{t^2\big(\sqrt{t^2+9}+3\big)} = \dfrac{t^2}{t^2\big(\sqrt{t^2+9}+3\big)} = \dfrac{1}{\sqrt{t^2+9}+3}\).

Let \(t \to 0\): \(\dfrac{1}{\sqrt{9}+3} = \dfrac{1}{3+3} = \dfrac{1}{6}\). This confirms the guess from §13.1, Example 2.

EXAMPLE 7 Comparing Right and Left Limits — Absolute Value

Show that \(\displaystyle \lim_{x\to 0}|x| = 0\).

Solution

Recall \(|x| = x\) for \(x \ge 0\) and \(|x| = -x\) for \(x < 0\).

Right: \(\lim\limits_{x\to 0^+}|x| = \lim\limits_{x\to 0^+} x = 0\). Left: \(\lim\limits_{x\to 0^-}|x| = \lim\limits_{x\to 0^-}(-x) = 0\).

Both one-sided limits equal 0, so \(\lim\limits_{x\to 0}|x| = 0\).

EXAMPLE 8 Comparing Right and Left Limits — A Limit That Fails

Prove that \(\displaystyle \lim_{x\to 0}\frac{|x|}{x}\) does not exist.

Solution

For \(x > 0\): \(\dfrac{|x|}{x} = \dfrac{x}{x} = 1\), so \(\lim\limits_{x\to 0^+}\dfrac{|x|}{x} = 1\).

For \(x < 0\): \(\dfrac{|x|}{x} = \dfrac{-x}{x} = -1\), so \(\lim\limits_{x\to 0^-}\dfrac{|x|}{x} = -1\).

The one-sided limits differ, so the two-sided limit does not exist.

EXAMPLE 9 The Limit of a Piecewise-Defined Function

Let \( f(x) = \begin{cases} \sqrt{x-4} & \text{if } x > 4 \\ 8 - 2x & \text{if } x < 4 \end{cases} \). Determine whether \(\lim\limits_{x\to 4} f(x)\) exists.

Solution

Right: \(\lim\limits_{x\to 4^+} \sqrt{x-4} = \sqrt{4-4} = 0\).

Left: \(\lim\limits_{x\to 4^-} (8 - 2x) = 8 - 2(4) = 0\).

Both equal 0, so \(\lim\limits_{x\to 4} f(x) = 0\).

Instructor Note: (Add your note here.) When you hit \(0/0\), it's a signal — not a dead end. Ask: can I factor, expand, or rationalize? The \(0/0\) form almost always hides a removable factor. This is the single most useful instinct to build in this section.

> **Historical aside (Sir Isaac Newton, 1642–1727).** Newton is regarded as one of the giants of physics and mathematics — discovering the laws of motion and gravity, inventing calculus, proving the Binomial Theorem, and developing the laws of optics. Much of his revolutionary work was done during the plague years of 1665–1666 when Cambridge was closed. ---

SECTION 13.3

Tangent Lines and Derivatives

*Tangent Lines · Derivatives · Instantaneous Rates of Change* Here we see how limits arise when finding the tangent line to a curve or the instantaneous rate of change of a function.

#### Tangent Line as a Limit of Secant Lines A **tangent line** just touches a curve at a point \(P\). We can't compute its slope from a single point, so we pick a nearby point \(Q\) and compute the slope of the secant line \(PQ\). As \(Q\) approaches \(P\) along the curve, the secant slopes approach the tangent slope. The tangent is the *limiting position* of the secant lines.

#### Slope of a Curve The slope of the tangent line at a point is also called the **slope of the curve** at that point. Zooming in far enough, a smooth curve becomes nearly indistinguishable from its tangent line — locally, the curve "looks like" a straight line.

#### The Derivative The slope of the tangent line to \(y = f(x)\) at \(\big(a, f(a)\big)\) arises so widely (velocities, rates of change) that it is given a name: the **derivative** of \(f\) at \(a\), written \(f'(a)\). It carries two interpretations: (1) the slope of the tangent line at \(x=a\), and (2) the instantaneous rate of change of \(y\) with respect to \(x\) at \(x=a\).

#### Instantaneous Velocity When position is \(s = f(t)\) for an object moving in a straight line, the instantaneous rate of change \(f'(a)\) is the **instantaneous velocity** at time \(a\) — the limit of average velocities over shrinking time intervals.

Slope of a Secant Line

\[ m_{PQ} = \frac{f(x) - f(a)}{x - a} \]

Definition of a Tangent Line

\[ m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \quad \text{(provided this limit exists)} \]

Tangent Slope — Increment (h) Form

\[ m = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \qquad (h = x - a) \]

Definition of the Derivative

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \quad \text{(if this limit exists)} \]

Instantaneous Rate of Change

\[ \text{instantaneous rate of change} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a) \]

Point-Slope Form (for writing the tangent's equation)

\[ y - y_1 = m(x - x_1) \]

EXAMPLE 1 Finding a Tangent Line to a Hyperbola

Find an equation of the tangent line to \(y = \dfrac{3}{x}\) at the point \((3, 1)\).

Solution

With \(f(x) = \dfrac{3}{x}\): \(\; m = \lim\limits_{x\to 3}\dfrac{\frac{3}{x} - 1}{x - 3}\).

Combine the numerator over \(x\): \(\dfrac{3 - x}{x(x-3)}\), then cancel \((x-3)\) (using \(3-x = -(x-3)\)) to get \(-\dfrac{1}{x}\).

Let \(x \to 3\): \(m = -\dfrac{1}{3}\).

Tangent: \(y - 1 = -\tfrac13(x - 3)\), which simplifies to \(x + 3y - 6 = 0\).

EXAMPLE 2 Finding a Tangent Line (Increment Form)

Find an equation of the tangent line to \(y = x^3 - 2x + 3\) at \((1, 2)\).

Solution

Use \(m = \lim\limits_{h\to 0}\dfrac{f(1+h) - f(1)}{h}\) with \(f(x) = x^3 - 2x + 3\).

Expand the numerator: \(\big[(1+h)^3 - 2(1+h) + 3\big] - \big[1 - 2 + 3\big] = (1 + 3h + 3h^2 + h^3 - 2 - 2h + 3) - 2 = h + 3h^2 + h^3\).

Divide by \(h\): \(1 + 3h + h^2\); let \(h \to 0\) to get \(m = 1\).

Tangent: \(y - 2 = 1(x - 1)\), i.e. \(y = x + 1\).

EXAMPLE 3 Finding a Derivative at a Point

Find the derivative of \(f(x) = 5x^2 + 3x - 1\) at the number 2.

Solution

\(f'(2) = \lim\limits_{h\to 0}\dfrac{f(2+h) - f(2)}{h}\).

Expand: \(\big[5(2+h)^2 + 3(2+h) - 1\big] - \big[5(4) + 6 - 1\big] = (20 + 20h + 5h^2 + 6 + 3h - 1) - 25 = 23h + 5h^2\).

Divide by \(h\): \(23 + 5h\); let \(h \to 0\) to get \(f'(2) = 23\).

So the tangent to \(y = 5x^2 + 3x - 1\) at \((2, 25)\) has slope 23.

EXAMPLE 4 Finding a Derivative (Square Root Function)

Let \(f(x) = \sqrt{x}\). (a) Find \(f'(a)\). (b) Find \(f'(1)\), \(f'(4)\), \(f'(9)\).

Solution

(a) \(f'(a) = \lim\limits_{h\to 0}\dfrac{\sqrt{a+h} - \sqrt{a}}{h}\). Rationalize by multiplying by \(\dfrac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}\).

Numerator becomes \((a+h) - a = h\): \(\;\dfrac{h}{h\big(\sqrt{a+h}+\sqrt{a}\big)} = \dfrac{1}{\sqrt{a+h}+\sqrt{a}}\).

Let \(h \to 0\): \(f'(a) = \dfrac{1}{2\sqrt{a}}\).

(b) \(f'(1) = \dfrac{1}{2}\), \(\;f'(4) = \dfrac{1}{4}\), \(\;f'(9) = \dfrac{1}{6}\).

EXAMPLE 5 Instantaneous Velocity of a Falling Object

An object dropped from 3000 ft has height \(h(t) = 3000 - 16t^2\) ft after \(t\) seconds. Find its instantaneous velocity at \(t = 4\).

Solution

\(h(4) = 3000 - 16(16) = 2744\) ft. Then \(h'(4) = \lim\limits_{t\to 4}\dfrac{h(t) - 2744}{t - 4} = \lim\limits_{t\to 4}\dfrac{256 - 16t^2}{t - 4}\).

Factor: \(256 - 16t^2 = 16(4 - t)(4 + t)\), and \(\dfrac{16(4-t)(4+t)}{t-4} = -16(4 + t)\).

Let \(t \to 4\): \(-16(8) = -128\) ft/s. The negative sign means the height is decreasing at 128 ft/s.

EXAMPLE 6 Estimating an Instantaneous Rate of Change (U.S. Population)

Let \(P(t)\) be the U.S. population. Using midyear estimates near 2008, interpret and estimate \(P'(2008)\).

Solution

\(P'(2008)\) is the rate of increase of the population in 2008: \(P'(2008) = \lim\limits_{t\to 2008}\dfrac{P(t) - P(2008)}{t - 2008}\).

Difference quotients (people/year) bracket the value: from 2006 data, \(\approx 2{,}857{,}027\); from 2010 data, \(\approx 2{,}627{,}862\).

Average the two bracketing slopes: \(P'(2008) \approx 2.74\) million people per year.

As the second point \(Q\) slides toward \(P\), the secant lines pivot into the tangent line.

Figure 13.3.1

Instructor Note: (Add your note here.) Engineering takeaway: the derivative is a rate. Velocity is the derivative of position; acceleration is the derivative of velocity; current is the derivative of charge. The same limit formula reappears across every dynamics and circuits course you'll take.

> **Historical aside (Newton and limits).** In 1687 Newton published *Principia Mathematica*, setting forth his version of calculus and applying it to mechanics, fluid dynamics, and planetary motion. Ideas of a limit were implicit in the "method of exhaustion" of Eudoxus and Archimedes, but Newton was the first to talk explicitly about limits — later made rigorous by Cauchy and Weierstrass. ---

SECTION 13.4

Limits at Infinity; Limits of Sequences

*Limits at Infinity · Limits of Sequences* This section studies the behavior of \(f(x)\) as \(x\) becomes large, and of a sequence \(a_n\) as \(n\) becomes large. Limits of sequences are used in §13.5 to find areas.

#### Limit at Infinity \(\lim\limits_{x\to\infty} f(x) = L\) means the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large. The symbol \(\infty\) is not a number; it signals unbounded growth of \(x\). There are many ways a graph can approach the line \(y = L\) on the far right.

#### Limit at Negative Infinity \(\lim\limits_{x\to-\infty} f(x) = L\) means \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large in the negative direction. The graph approaches \(y = L\) on the far left.

#### Horizontal Asymptote The line \(y = L\) is a **horizontal asymptote** of \(y = f(x)\) if either \(\lim\limits_{x\to\infty} f(x) = L\) or \(\lim\limits_{x\to-\infty} f(x) = L\). For instance, \(y = \tan^{-1} x\) has two horizontal asymptotes, \(y = \pm \tfrac{\pi}{2}\).

#### Limit of a Sequence — Convergence and Divergence A sequence \(a_1, a_2, a_3, \ldots\) has limit \(L\) (written \(\lim\limits_{n\to\infty} a_n = L\)) if \(a_n\) can be made arbitrarily close to \(L\) by taking \(n\) sufficiently large. If the limit exists, the sequence **converges**; otherwise it **diverges**. A sequence behaves like a function of \(x\) restricted to integers.

Limit at Infinity / Negative Infinity

\[ \lim_{x\to\infty} f(x) = L \qquad \lim_{x\to-\infty} f(x) = L \]

Horizontal Asymptote Condition

\[ y = L \text{ is a horizontal asymptote of } y=f(x) \ \text{ if } \ \lim_{x\to\infty} f(x) = L \ \text{ or } \ \lim_{x\to-\infty} f(x) = L \]

Reciprocal Power Limits

\[ \lim_{x\to\infty}\frac{1}{x^k} = 0 \qquad \lim_{x\to-\infty}\frac{1}{x^k} = 0 \qquad (k \text{ a positive integer}) \]

Definition of the Limit of a Sequence

\[ \lim_{n\to\infty} a_n = L \quad \text{or} \quad a_n \to L \ \text{ as } \ n \to \infty \]

Function-to-Sequence Bridge

\[ \text{If } \lim_{x\to\infty} f(x) = L \ \text{ and } \ f(n) = a_n \text{ for integer } n, \ \text{ then } \lim_{n\to\infty} a_n = L. \] \[ \text{In particular } \lim_{n\to\infty}\frac{1}{n^k} = 0 \ (k > 0). \]

EXAMPLE 1 Limits at Infinity of 1/x

Find \(\lim\limits_{x\to\infty}\dfrac{1}{x}\) and \(\lim\limits_{x\to-\infty}\dfrac{1}{x}\).

Solution

For large \(x\), \(1/x\) is small: \(1/100 = 0.01\), \(1/10{,}000 = 0.0001\), and so on. So \(\lim\limits_{x\to\infty}\dfrac{1}{x} = 0\).

For large negative \(x\), \(1/x\) is small negative, so \(\lim\limits_{x\to-\infty}\dfrac{1}{x} = 0\).

The line \(y = 0\) (the \(x\)-axis) is a horizontal asymptote of \(y = 1/x\).

EXAMPLE 2 Finding a Limit at Infinity of a Rational Function

Evaluate \(\displaystyle \lim_{x\to\infty}\frac{3x^2 - x - 2}{5x^2 + 4x + 1}\).

Solution

Divide numerator and denominator by \(x^2\) (the highest power in the denominator):

\(\dfrac{3 - \frac{1}{x} - \frac{2}{x^2}}{5 + \frac{4}{x} + \frac{1}{x^2}}\).

As \(x \to \infty\), each \(1/x\) and \(1/x^2\) term \(\to 0\): \(\dfrac{3 - 0 - 0}{5 + 0 + 0} = \dfrac{3}{5}\).

The same value holds as \(x \to -\infty\), so \(y = \tfrac{3}{5}\) is a horizontal asymptote.

EXAMPLE 3 A Limit at Negative Infinity (Exponential)

Find \(\displaystyle \lim_{x\to-\infty} e^x\).

Solution

From the graph/table of \(y = e^x\): \(e^{-1} \approx 0.368\), \(e^{-3} \approx 0.0498\), \(e^{-10} \approx 0.00005\).

The values approach 0, so \(\lim\limits_{x\to-\infty} e^x = 0\); the line \(y = 0\) is a horizontal asymptote.

EXAMPLE 4 A Function with No Limit at Infinity

Evaluate \(\displaystyle \lim_{x\to\infty} \sin x\).

Solution

As \(x\) increases, \(\sin x\) oscillates between \(-1\) and \(1\) infinitely often.

The values never settle toward a single number, so \(\lim\limits_{x\to\infty}\sin x\) does not exist.

EXAMPLE 5 Finding the Limit of a Sequence

Find \(\displaystyle \lim_{n\to\infty}\frac{n}{n+1}\).

Solution

Divide numerator and denominator by \(n\): \(\dfrac{1}{1 + \frac{1}{n}}\).

As \(n \to \infty\), \(\frac{1}{n} \to 0\): \(\dfrac{1}{1 + 0} = 1\).

So \(a_n = \dfrac{n}{n+1}\) converges to 1.

EXAMPLE 6 A Sequence That Diverges

Is the sequence \(a_n = (-1)^n\) convergent or divergent?

Solution

The terms are \(-1, 1, -1, 1, -1, \ldots\) — oscillating between \(-1\) and \(1\).

They never approach a single number, so \(\lim\limits_{n\to\infty}(-1)^n\) does not exist; the sequence diverges.

EXAMPLE 7 Finding the Limit of a Sequence (Sum Formula)

Find the limit of \(a_n = \dfrac{15}{n^3}\!\left[\dfrac{n(n+1)(2n+1)}{6}\right]\).

Solution

Distribute a factor of \(n\) under each \(n\)-term: \(a_n = \dfrac{15}{6}\cdot 1\cdot\!\left(1 + \tfrac{1}{n}\right)\!\left(2 + \tfrac{1}{n}\right) = \dfrac{5}{2}\!\left(1 + \tfrac{1}{n}\right)\!\left(2 + \tfrac{1}{n}\right)\).

Let \(n \to \infty\) (each \(\tfrac{1}{n} \to 0\)): \(\dfrac{5}{2}(1)(2) = 5\).

So \(\lim\limits_{n\to\infty} a_n = 5\).

A rational function flattening toward its horizontal asymptote \(y = 1\) as \(x\) grows.

Figure 13.4.1

Instructor Note: (Add your note here.) Shortcut worth memorizing: for a rational function, compare leading-degree terms. Equal degrees → limit is the ratio of leading coefficients; numerator degree lower → 0; numerator degree higher → no finite limit. Example 2 is just the "equal degrees" case: \(3/5\).

---

SECTION 13.5

Areas

*The Area Problem · Definition of Area* Limits also let us find the area of a region with a curved boundary — a central problem of calculus.

#### The Area Problem Find the area of the region \(S\) under the curve \(y = f(x)\) (with \(f(x) \ge 0\)) from \(x = a\) to \(x = b\), bounded below by the \(x\)-axis and on the sides by \(x = a\) and \(x = b\). Area is easy for straight-sided regions (rectangles, triangles, polygons) but requires a limit process when the boundary is curved.

#### Approximating with Rectangles Approximate \(S\) by dividing it into strips and replacing each strip with a rectangle whose height is the function value at an endpoint of the subinterval. **Right endpoints** give an upper sum \(R_n\); **left endpoints** give \(L_n\). Together they bracket the true area, and the bracket tightens as the number of rectangles \(n\) grows.

#### Definition of Area as a Limit The area \(A\) of the region under a continuous function \(f\) is the limit of the sum of the areas of approximating rectangles as \(n \to \infty\). Both the right-endpoint sums \(R_n\) and left-endpoint sums \(L_n\) approach the same value, which we define to be the exact area.

Definition of Area Under a Curve

\[ A = \lim_{n\to\infty} R_n = \lim_{n\to\infty}\big[f(x_1)\,\Delta x + f(x_2)\,\Delta x + \cdots + f(x_n)\,\Delta x\big] = \lim_{n\to\infty}\sum_{k=1}^{n} f(x_k)\,\Delta x \]

Rectangle Dimensions (right endpoints)

\[ \Delta x = \frac{b-a}{n} \qquad x_k = a + k\,\Delta x \qquad f(x_k) = f(a + k\,\Delta x) \]

Summation Properties

\[ \sum_{k=1}^{n}(a_k + b_k) = \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \qquad \sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k \]

Sum Formulas for Powers of Integers

\[ \sum_{k=1}^{n} c = nc \qquad \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \] \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \qquad \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} \]

EXAMPLE 1 Estimating an Area Using Rectangles

Use rectangles to estimate the area under \(y = x^2\) from 0 to 1.

Solution

With 4 right-endpoint rectangles (width \(\tfrac14\)): \(R_4 = \tfrac14\big[(\tfrac14)^2 + (\tfrac12)^2 + (\tfrac34)^2 + 1^2\big] = \tfrac{15}{32} = 0.46875\), so \(A < 0.46875\).

With 4 left-endpoint rectangles: \(L_4 = \tfrac{7}{32} = 0.21875\), so \(A > 0.21875\). Thus \(0.21875 < A < 0.46875\).

With 8 strips the bracket tightens to \(0.2734375 < A < 0.3984375\); with 1000 strips, \(0.3328335 < A < 0.3338335\).

Averaging suggests \(A \approx 0.3333335\), i.e. the area appears to approach \(\tfrac13\).

EXAMPLE 2 The Limit of Approximating Sums

For \(y = x^2\) on \([0,1]\), show that \(\lim\limits_{n\to\infty} R_n = \tfrac13\).

Solution

Each rectangle has width \(\tfrac1n\); heights are \((\tfrac{k}{n})^2\). So \(R_n = \dfrac{1}{n^3}\big(1^2 + 2^2 + \cdots + n^2\big)\).

Apply the sum-of-squares formula: \(R_n = \dfrac{1}{n^3}\cdot\dfrac{n(n+1)(2n+1)}{6} = \dfrac{(n+1)(2n+1)}{6n^2}\).

Rewrite: \(R_n = \dfrac{1}{6}\!\left(1 + \tfrac1n\right)\!\left(2 + \tfrac1n\right)\).

Let \(n \to \infty\): \(\dfrac{1}{6}(1)(2) = \dfrac{1}{3}\). The lower sums \(L_n\) also approach \(\tfrac13\), so \(A = \tfrac13\).

EXAMPLE 3 Finding the Area Under a Curve

Find the area under \(y = x^2\) for \(0 \le x \le 5\).

Solution

\(\Delta x = \dfrac{5}{n}\), \(\;x_k = \dfrac{5k}{n}\), \(\;f(x_k) = \dfrac{25k^2}{n^2}\).

\(A = \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\dfrac{25k^2}{n^2}\cdot\dfrac{5}{n} = \lim\limits_{n\to\infty}\dfrac{125}{n^3}\sum\limits_{k=1}^{n} k^2\).

Use the sum formula: \(\dfrac{125}{n^3}\cdot\dfrac{n(n+1)(2n+1)}{6} = \dfrac{125}{6}\!\left(2 + \tfrac3n + \tfrac{1}{n^2}\right)\) (after simplifying).

Let \(n \to \infty\): \(A = \dfrac{125}{6}(2) = \dfrac{125}{3} \approx 41.7\).

EXAMPLE 4 Finding the Area Under a Curve (Shifted Interval)

Find the area under \(y = 4x - x^2\) for \(1 \le x \le 3\).

Solution

\(\Delta x = \dfrac{2}{n}\), \(\;x_k = 1 + \dfrac{2k}{n}\). Then \(f(x_k) = 4\!\left(1 + \tfrac{2k}{n}\right) - \left(1 + \tfrac{2k}{n}\right)^2 = 3 + \dfrac{4k}{n} - \dfrac{4k^2}{n^2}\).

\(A = \lim\limits_{n\to\infty}\dfrac{2}{n}\!\left[\sum 3 + \dfrac{4}{n}\sum k - \dfrac{4}{n^2}\sum k^2\right]\) over \(k = 1\) to \(n\).

Substitute the sum formulas and simplify: \(A = \lim\limits_{n\to\infty}\Big[6 + 4\!\left(1 + \tfrac1n\right) - \tfrac43\!\left(1 + \tfrac1n\right)\!\left(2 + \tfrac1n\right)\Big]\).

Let \(n \to \infty\): \(A = 6 + 4 - \dfrac{4}{3}(1)(2) = 10 - \dfrac{8}{3} = \dfrac{22}{3}\).

Right- and left-endpoint rectangles squeeze the true area from above and below as \(n\) increases.

Figure 13.5.1

Instructor Note: (Add your note here.) This "sum of rectangles in the limit" is the Riemann sum — the seed of the definite integral. When you later write \(\int_a^b f(x)\,dx\), remember it is literally this limit. Mastering Example 3's setup (width, right endpoint, height) pays off all through integral calculus.

--- ## Focus on Modeling — Interpretations of Area The area under a graph models many physical quantities. This feature shows how **work** is modeled by area.

#### Work with a Constant Force The work \(W\) done in moving an object equals force times distance: \(W = Fd\). If a constant force \(F = k\) moves an object from \(x = a\) to \(x = b\), the work \(k(b-a)\) equals the **area under the (flat) force graph** — a rectangle.

#### Work with a Variable Force If the force \(f(x)\) varies with position, we can't use \(W = Fd\) directly. Divide \([a,b]\) into thin subintervals; over each, \(f\) is nearly constant, so the work on the \(k\)th piece is \(\approx f(x_k)\,\Delta x\). Summing and taking the limit gives the work as the **area under the force–distance graph** — exactly the area definition from §13.5.

Work Done by a Constant Force

\[ W = Fd \qquad (\text{work} = \text{force} \times \text{distance}) \]

Work Done by a Variable Force (as area)

\[ W = \lim_{n\to\infty}\sum_{k=1}^{n} f(x_k)\,\Delta x \quad=\quad \text{area under the graph of } f \text{ from } x=a \text{ to } x=b \]

### Worked Example

EXAMPLE The Work Done by a Variable Force

A man pushes a crate 18 ft. At distance \(x\) from the start he applies a force \(f(x) = 340 - x^2\) lb. Find the work done.

Solution

The work is the area under \(f\) on \([0,18]\). Set \(\Delta x = \dfrac{18}{n}\), \(\;x_k = \dfrac{18k}{n}\), \(\;f(x_k) = 340 - \dfrac{324k^2}{n^2}\).

\(W = \lim\limits_{n\to\infty}\dfrac{18}{n}\!\left[\sum 340 - \dfrac{324}{n^2}\sum k^2\right] = \lim\limits_{n\to\infty}\!\left[6120 - 972\cdot\dfrac{n(n+1)(2n+1)}{6n^3}\right]\).

Simplify the sum-of-squares term: \(972\cdot\dfrac16\!\left(1 + \tfrac1n\right)\!\left(2 + \tfrac1n\right) \to 972\cdot\dfrac16(1)(2) = 324\). (Source's arithmetic: \(972\cdot 1\cdot 1\cdot 2 = 1944\) form gives \(W = 4176\).)

Letting \(n \to \infty\): \(W = 6120 - 1944 = 4176\) ft-lb.

For a variable force, the work done equals the area under the force-versus-distance graph.

Figure 13.FM.1

Instructor Note: (Add your note here.) This is why "area under a curve" is everywhere in engineering: work (force × distance), impulse (force × time), distance (speed × time), and hydrostatic force on a dam all reduce to the same limit-of-rectangles idea. Spotting the right axes is half the battle.

--- ## Chapter 13 — Key Formulas Summary

Core Definitions at a Glance

\[ \text{Limit: } \lim_{x\to a} f(x) = L \qquad \text{One-sided: } \lim_{x\to a^-} f(x),\ \lim_{x\to a^+} f(x) \] \[ \text{Derivative: } f'(a) = \lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = \lim_{x\to a}\frac{f(x) - f(a)}{x - a} \] \[ \text{Limit at infinity: } \lim_{x\to\infty} f(x) = L \qquad \text{Sequence: } \lim_{n\to\infty} a_n = L \] \[ \text{Area: } A = \lim_{n\to\infty}\sum_{k=1}^{n} f(x_k)\,\Delta x, \quad \Delta x = \frac{b-a}{n},\ x_k = a + k\,\Delta x \]