Vector-Valued Functions and Motion: Background You’ll Need 1
Use basic rules to find limits of functions, including polynomial and rational functions
The Limit Laws
Limit laws are algebraic rules that let us break down complicated limits into simpler parts. Instead of evaluating limits from scratch each time, we can use these properties to handle complex functions systematically.
basic limit properties
Let [latex]a[/latex] be a real number and [latex]c[/latex] be a constant.
[latex]\underset{x\to a}{\lim}x=a[/latex]
[latex]\underset{x\to a}{\lim}c=c[/latex]
Evaluate each of the following limits using the basic limit properties above.
[latex]\underset{x\to 2}{\lim}x[/latex]
[latex]\underset{x\to 2}{\lim}5[/latex]
The limit of [latex]x[/latex] as [latex]x[/latex] approaches [latex]a[/latex] is [latex]a[/latex]: [latex]\underset{x\to 2}{\lim}x=2[/latex].
The limit of a constant is that constant: [latex]\underset{x\to 2}{\lim}5=5[/latex].
The limit laws outline the essential properties of limits, crucial for the systematic evaluation of functions as they approach specific points.
the limit laws
For all [latex]x[/latex] near [latex]a[/latex], consider functions [latex]f(x)[/latex] and [latex]g(x)[/latex] with limits [latex]L[/latex] and [latex]M[/latex] respectively, [latex]\underset{x\to a}{\lim}f(x)=L[/latex] and [latex]\underset{x\to a}{\lim}g(x)=M[/latex]. Let [latex]c[/latex] be a constant. The following are established limit laws:
[latex]\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}[/latex] for [latex]M\ne 0[/latex]
Power law for limits:
[latex]\underset{x\to a}{\lim}(f(x))^n=(\underset{x\to a}{\lim}f(x))^n=L^n[/latex] for every positive integer [latex]n[/latex]
.
Root law for limits:
[latex]\underset{x\to a}{\lim}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\lim}f(x)}=\sqrt[n]{L}[/latex] for all [latex]L[/latex] if [latex]n[/latex] is odd and for [latex]L\ge 0[/latex] if [latex]n[/latex] 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 [latex]\underset{x\to -3}{\lim}(4x+2)[/latex].
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.
[latex]\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}[/latex]
Use the limit laws to evaluate [latex]\underset{x\to 2}{\lim}\dfrac{2x^2-3x+1}{x^3+4}[/latex].
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.
[latex]\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}[/latex]
Limits of Polynomial and Rational Functions
When working with polynomial and rational functions, you can often find limits by direct substitution.
Consider the limit [latex]\underset{x\to 2}{\text{lim}}\dfrac{2x^2-3x+1}{x^3+4}[/latex]. Using limit laws step-by-step gives us [latex]\frac{1}{4}[/latex]. But notice what happens if we simply substitute [latex]x = 2[/latex] directly into the function:
This is not mere chance. It demonstrates a foundational concept in calculus: for polynomial and rational functions that are continuous at a point [latex]a[/latex], the limit as [latex]x[/latex] approaches [latex]a[/latex] equals the value of the function at [latex]a[/latex], or [latex]f(a)[/latex]. This holds true provided the function is defined at that point.
limits of polynomial and rational functions
Let [latex]p(x)[/latex] and [latex]q(x)[/latex] be polynomial functions. Let [latex]a[/latex] be a real number. Then,
Evaluate the [latex]\underset{x\to 3}{\lim}\dfrac{2x^2-3x+1}{5x+4}[/latex].
Since [latex]3[/latex] is in the domain of the rational function [latex]f(x)=\frac{2x^2-3x+1}{5x+4}[/latex], we can calculate the limit by substituting [latex]3[/latex] for [latex]x[/latex] into the function.