The Limit Laws: Learn It 1

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The Limit Laws: Learn It 1

Use basic rules to find limits for polynomial and rational functions
Find function limits by breaking them into simpler parts (factoring) or using conjugates
Determine limits by applying the squeeze theorem

Evaluating Limits

The Limit Laws

As we continue our journey through calculus, we encounter limit laws—critical tools that help us understand how functions behave as inputs approach a certain value. These laws are fundamental for calculating function limits and serve as a gateway to deeper concepts like continuity and differentiation.

Through our initial introduction to limits, we identified two properties that are particularly significant. These properties, along with other limit laws, allow us to evaluate limits for a wide variety of algebraic functions with ease.

basic limit properties

Let $a$ be a real number and $c$ be a constant.

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

Evaluate each of the following limits using the basic limit properties above.

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

The limit of $x$ as $x$ approaches $a$ is $a$ : $\underset{x\to 2}{\lim}x=2$ .
The limit of a constant is that constant: $\underset{x\to 2}{\lim}5=5$ .

The limit laws outline the essential properties of limits, crucial for the systematic evaluation of functions as they approach specific points. Our focus will be on their practical use, since the detailed proofs are beyond this course’s scope.

the limit laws

For all $x$ near $a$ , consider functions $f(x)$ and $g(x)$ with limits $L$ and $M$ respectively, $\underset{x\to a}{\lim}f(x)=L$ and $\underset{x\to a}{\lim}g(x)=M$ . Let $c$ be a constant. The following are established limit laws:

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

$\underset{x\to a}{\lim}(f(x)+g(x))=\underset{x\to a}{\lim}f(x)+\underset{x\to a}{\lim}g(x)=L+M$

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

$\underset{x\to a}{\lim}(f(x)-g(x))=\underset{x\to a}{\lim}f(x)-\underset{x\to a}{\lim}g(x)=L-M$

Constant multiple law for limits:

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

$\underset{x\to a}{\lim}cf(x)=c \cdot \underset{x\to a}{\lim}f(x)=cL$

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

$\underset{x\to a}{\lim}(f(x) \cdot g(x))=\underset{x\to a}{\lim}f(x) \cdot \underset{x\to a}{\lim}g(x)=L \cdot M$

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}

$\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\dfrac{\underset{x\to a}{\lim}f(x)}{\underset{x\to a}{\lim}g(x)}=\frac{L}{M}$ for

M \neq 0

$M\ne 0$

Power law for limits:

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

$\underset{x\to a}{\lim}(f(x))^n=(\underset{x\to a}{\lim}f(x))^n=L^n$ for every positive integer

n

$n$

.

Root law for limits:

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

$\underset{x\to a}{\lim}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\lim}f(x)}=\sqrt[n]{L}$ for all

L

$L$ if

n

$n$ is odd and for

L \geq 0

$L\ge 0$ if

n

$n$ is even

A handy way to keep the limit laws top of mind is to associate them with simple arithmetic operations you already know:

Sum and Difference: Just like adding or subtracting numbers, you can add or subtract limits.
Constant Multiples: Multiplying a number by a constant? The same goes for a limit.
Product: Multiplying two numbers? You can also multiply their limits.
Quotient: Dividing numbers translates to dividing their limits, just watch out for a zero denominator.
Powers and Roots: Raising a number to a power or taking a root? Apply the same operation to the limit.

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

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

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

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

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

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

Limits of Polynomial and Rational Functions

When exploring limits of polynomial and rational functions, a notable pattern emerges.

Let’s examine the recent example where we calculated the limit of $\dfrac{2x^2-3x+1}{x^3+4}$ as $x$ approaches $2$ . Applying the limit laws, the limit was found to be $\frac{1}{4}$ . If $x$ is replaced with $2$ in the function directly, $f(2)=\dfrac{2(2)^2-3(2)+1}{(2)^3+4}$ , $f(2)$ is also equal to $\frac{1}{4}$ .

This is not mere chance. It demonstrates a foundational concept in calculus: for polynomial and rational functions that are continuous at a point $a$ , the limit as $x$ approaches $a$ equals the value of the function at $a$ , or $f(a)$ . This holds true provided the function is defined at that point.

limits of polynomial and rational functions

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

lim_{x \to a} p (x) = p (a)

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

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)} when q (a) \neq 0

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

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 power laws, we end up with:

\begin{array}{cc} lim_{x \to a} p (x) & = lim_{x \to a} (c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0}) \\ = c_{n} (lim_{x \to a} x)^{n} + c_{n - 1} (lim_{x \to a} x)^{n - 1} + \dots + c_{1} (lim_{x \to a} x) + lim_{x \to a} c_{0} \\ = c_{n} a^{n} + c_{n - 1} a^{n - 1} + \dots + c_{1} a + c_{0} \\ = p (a) \end{array}

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

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

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)}

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

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

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

$f(3)=\frac{2(3)^2-3(3)+1}{5(3)+4} = \dfrac{10}{19}$

Thus,

lim_{x \to 3} \frac{2 x^{2} - 3 x + 1}{5 x + 4} = \frac{10}{19}

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