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Limits

Sarah Miller

Sarah Miller

7 min read

Next Topic - Evaluating Limits Analytically

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Study Guide Overview

This study guide covers properties of limits, including limits of constant functions, sums, differences, products, quotients, powers, and composite functions. It also explores infinite limits and how limit properties apply to them. The guide provides worked examples, practice questions, a glossary of key terms, and exam tips for applying these properties effectively.

#Properties of Limits

#Table of Contents

  1. Introduction to Limit Properties
  2. Basic Limit Properties
    • Limit of a Constant Function
    • Limit of a Multiple of a Function
    • Limit of a Sum or Difference of Functions
    • Limit of a Product of Functions
    • Limit of a Quotient of Functions
    • Limit of the Power of a Function
    • Limit of a Composite Function
  3. Infinite Limits
  4. Practice Questions
  5. Glossary
  6. Summary and Key Takeaways

#Introduction to Limit Properties

Understanding limit properties (also known as limit theorems) is crucial for solving complex functions algebraically. These properties allow us to break down complicated functions into simpler parts, making it easier to determine their limits.

#Basic Limit Properties

#Limit of a Constant Function

If kkk is a constant, then: lim⁡x→ak=k\lim_{{x \to a}} k = kx→alim​k=k

#Limit of a Multiple of a Function

If kkk is a constant and lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L, then: lim⁡x→a(kf(x))=kL\lim_{{x \to a}} (k f(x)) = kLx→alim​(kf(x))=kL

#Limit of a Sum or Difference of Functions

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and lim⁡x→ag(x)=M\lim_{{x \to a}} g(x) = Mlimx→a​g(x)=M, then: lim⁡x→a(f(x)±g(x))=L±M\lim_{{x \to a}} (f(x) \pm g(x)) = L \pm Mx→alim​(f(x)±g(x))=L±M

#Limit of a Product of Functions

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and lim⁡x→ag(x)=M\lim_{{x \to a}} g(x) = Mlimx→a​g(x)=M, then: lim⁡x→a(f(x)⋅g(x))=L⋅M\lim_{{x \to a}} (f(x) \cdot g(x)) = L \cdot Mx→alim​(f(x)⋅g(x))=L⋅M

#Limit of a Quotient of Functions

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and lim⁡x→ag(x)=M\lim_{{x \to a}} g(x) = Mlimx→a​g(x)=M with M≠0M \ne 0M=0, then: lim⁡x→af(x)g(x)=LM\lim_{{x \to a}} \frac{f(x)}{g(x)} = \frac{L}{M}x→alim​g(x)f(x)​=ML​

#Limit of the Power of a Function

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and nnn is a real number, then: lim⁡x→a[f(x)]n=Ln\lim_{{x \to a}} [f(x)]^n = L^nx→alim​[f(x)]n=Ln

#Limit of a Composite Function

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and the function ggg is continuous at x=Lx = Lx=L, then: lim⁡x→ag(f(x))=g(L)\lim_{{x \to a}} g(f(x)) = g(L)x→alim​g(f(x))=g(L)

Statements like lim⁡_x→af(x)=L\lim\_{{x \to a}} f(x) = Llim_x→af(x)=L and lim⁡_x→ag(x)=M\lim\_{{x \to a}} g(x) = Mlim_x→ag(x)=M assume that those limits exist and that LLL and MMM are real numbers.
Exam Tip

Make sure that the necessary conditions are met before using one of the limit properties:

  • M≠0M \ne 0M=0 for the quotient property
  • ggg is continuous at x=Lx = Lx=L for the composite function property
  • Both limits, if being combined, are tending toward the same value (a)(a)(a)

#Worked Example

Let fff and ggg be functions such that lim⁡x→3f(x)=7\lim_{{x \to 3}} f(x) = 7limx→3​f(x)=7 and lim⁡x→3g(x)=−2\lim_{{x \to 3}} g(x) = -2limx→3​g(x)=−2.

Let hhh be a function that is continuous for all real numbers, and such that h(−2)=0h(-2) = 0h(−2)=0 and h(7)=13h(7) = 13h(7)=13.

Find the following limits:

(a) lim⁡x→3(f(x)+4)\lim_{{x \to 3}} (f(x) + 4)limx→3​(f(x)+4)

Note that lim⁡_x→3(4)=4\lim\_{{x \to 3}} (4) = 4lim_x→3(4)=4, then use the limit of a sum of functions property:

lim⁡x→3(f(x)+4)=7+4\lim_{{x \to 3}} (f(x) + 4) = 7 + 4x→3lim​(f(x)+4)=7+4 lim⁡x→3(f(x)+4)=11\lim_{{x \to 3}} (f(x) + 4) = 11x→3lim​(f(x)+4)=11

(b) lim⁡x→3(g(x)−2f(x))\lim_{{x \to 3}} (g(x) - 2f(x))limx→3​(g(x)−2f(x))

Use the limit of a multiple of a function property, along with the limit of a difference of functions property:

lim⁡x→3(g(x)−2f(x))=−2−2(7)\lim_{{x \to 3}} (g(x) - 2f(x)) = -2 - 2(7)x→3lim​(g(x)−2f(x))=−2−2(7) lim⁡x→3(g(x)−2f(x))=−16\lim_{{x \to 3}} (g(x) - 2f(x)) = -16x→3lim​(g(x)−2f(x))=−16

(c) lim⁡x→3g(x)f(x)\lim_{{x \to 3}} \frac{g(x)}{f(x)}limx→3​f(x)g(x)​

Use the limit of a quotient of functions property:

lim⁡x→3g(x)f(x)=−27\lim_{{x \to 3}} \frac{g(x)}{f(x)} = \frac{-2}{7}x→3lim​f(x)g(x)​=7−2​ lim⁡x→3g(x)f(x)=−27\lim_{{x \to 3}} \frac{g(x)}{f(x)} = -\frac{2}{7}x→3lim​f(x)g(x)​=−72​

(d) lim⁡x→3h(f(x))\lim_{{x \to 3}} h(f(x))limx→3​h(f(x))

Use the limit of a composite function property. Note that hhh is continuous for all real numbers, so the property is valid for use here:

lim⁡x→3h(f(x))=h(7)\lim_{{x \to 3}} h(f(x)) = h(7)x→3lim​h(f(x))=h(7) lim⁡x→3h(f(x))=13\lim_{{x \to 3}} h(f(x)) = 13x→3lim​h(f(x))=13

#Infinite Limits

#How do the properties of limits work with infinite limits?

Several properties of limits involve infinite limits, helping to determine whether a function increases without bound (i.e., tends to ∞\infty∞) or decreases without bound (i.e., tends to −∞-\infty−∞) at a particular point.

#Limit of 1xn\frac{1}{x^n}xn1​ as xxx Approaches 0

If nnn is a positive integer, then: lim⁡x→0+1xn=∞\lim_{{x \to 0^+}} \frac{1}{x^n} = \inftyx→0+lim​xn1​=∞ lim⁡x→0−1xn={∞if n is even−∞if n is odd\lim_{{x \to 0^-}} \frac{1}{x^n} = \begin{cases} \infty & \text{if } n \text{ is even} \\ -\infty & \text{if } n \text{ is odd} \end{cases}x→0−lim​xn1​={∞−∞​if n is evenif n is odd​

#Infinite Limits of Quotients

If lim⁡x→af(x)=L\lim_{{x \to a}} f(x) = Llimx→a​f(x)=L and lim⁡x→ag(x)=0\lim_{{x \to a}} g(x) = 0limx→a​g(x)=0, then:

  • If L>0L > 0L>0, lim⁡x→af(x)g(x)={∞if g(x)>0 as x approaches a−∞if g(x)<0 as x approaches a\lim_{{x \to a}} \frac{f(x)}{g(x)} = \begin{cases} \infty & \text{if } g(x) > 0 \text{ as } x \text{ approaches } a \\ -\infty & \text{if } g(x) < 0 \text{ as } x \text{ approaches } a \end{cases}x→alim​g(x)f(x)​={∞−∞​if g(x)>0 as x approaches aif g(x)<0 as x approaches a​
  • If L<0L < 0L<0, lim⁡x→af(x)g(x)={−∞if g(x)>0 as x approaches a∞if g(x)<0 as x approaches a\lim_{{x \to a}} \frac{f(x)}{g(x)} = \begin{cases} -\infty & \text{if } g(x) > 0 \text{ as } x \text{ approaches } a \\ \infty & \text{if } g(x) < 0 \text{ as } x \text{ approaches } a \end{cases}x→alim​g(x)f(x)​={−∞∞​if g(x)>0 as x approaches aif g(x)<0 as x approaches a​

In both these cases, the limits from the left (as x→a−x \to a^-x→a−) and right (as x→a+x \to a^+x→a+) may be different. Check the behavior of g(x)g(x)g(x) to determine the correct limit.

#Worked Example

Let fff be a function such that lim⁡x→0f(x)=1\lim_{{x \to 0}} f(x) = 1limx→0​f(x)=1.

Let ggg be the function defined by g(x)=x3g(x) = x^3g(x)=x3.

Find lim⁡x→0−f(x)g(x)\lim_{{x \to 0^-}} \frac{f(x)}{g(x)}limx→0−​g(x)f(x)​ and lim⁡x→0+f(x)g(x)\lim_{{x \to 0^+}} \frac{f(x)}{g(x)}limx→0+​g(x)f(x)​.

We can use the infinite limits of quotient properties here. In both cases, lim⁡_x→0f(x)=L>0\lim\_{{x \to 0}} f(x) = L > 0lim_x→0f(x)=L>0.

For the limit from the left, note that g(x)=x3g(x) = x^3g(x)=x3 is negative as it approaches 0 through the negative numbers: lim⁡x→0−f(x)g(x)=−∞\lim_{{x \to 0^-}} \frac{f(x)}{g(x)} = -\inftyx→0−lim​g(x)f(x)​=−∞

For the limit from the right, note that g(x)=x3g(x) = x^3g(x)=x3 is positive as it approaches 0 through the positive numbers: lim⁡x→0+f(x)g(x)=∞\lim_{{x \to 0^+}} \frac{f(x)}{g(x)} = \inftyx→0+lim​g(x)f(x)​=∞

#Practice Questions

Practice Question

Question 1: Evaluate lim⁡x→2(3x2−4x+5)\lim_{{x \to 2}} (3x^2 - 4x + 5)limx→2​(3x2−4x+5).

Question 2: Find lim⁡x→−1x2−1x+1\lim_{{x \to -1}} \frac{x^2 - 1}{x + 1}limx→−1​x+1x2−1​.

Question 3: Determine lim⁡x→0sin⁡xx\lim_{{x \to 0}} \frac{\sin x}{x}limx→0​xsinx​.

Question 4: Evaluate lim⁡x→24x−1\lim_{{x \to 2}} \sqrt{4x - 1}limx→2​4x−1​.

#Glossary

  • Limit: The value that a function approaches as the input approaches some value.
  • Continuous Function: A function that is unbroken and smooth, having no gaps or jumps.
  • Quotient: The result of division.
  • Composite Function: A function that is formed when one function is substituted into another.

#Summary and Key Takeaways

  • The properties of limits allow us to simplify and solve complex functions algebraically by breaking them down into simpler parts.
  • Always ensure that the conditions for each limit property are met before applying them.
  • Understanding infinite limits is essential for dealing with functions that increase or decrease without bound.

#Key Takeaways

  • Basic Properties of Limits include limits of constants, sums, products, quotients, powers, and composite functions.
  • Infinite Limits help determine the behavior of functions as they tend to infinity or negative infinity.
  • Practice applying these limit properties to various functions to strengthen your understanding.

By mastering these properties, you will be better equipped to handle a wide range of limit problems in calculus.

Exam Tip

Always double-check the conditions before applying limit properties, especially for quotient and composite functions.

#Exam Strategy

  • Read the Problem Carefully: Ensure you understand what is being asked before starting any calculations.
  • Check Conditions: Verify that all necessary conditions for using limit properties are met.
  • Simplify Step-by-Step: Break down complex functions into simpler parts using the limit properties.
  • Practice: Regular practice with different types of limit problems will enhance your problem-solving skills.

By following these strategies, you can approach limit problems with confidence and accuracy.

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