Limit Laws

You've studied the definition of limits and how to find them algebraically and graphically. Is there perhaps a faster way to find the limit of a function. Yes, there is! We can use limit laws. Here you will see some of the more common properties of limits of functions and how to apply them.

Get started

Millions of flashcards designed to help you ace your studies

Sign up for free

Need help?
Meet our AI Assistant

Upload Icon

Create flashcards automatically from your own documents.

   Upload Documents
Upload Dots

FC Phone Screen

Need help with
Limit Laws?
Ask our AI Assistant

Review generated flashcards

Sign up for free
You have reached the daily AI limit

Start learning or create your own AI flashcards

StudySmarter Editorial Team

Team Limit Laws Teachers

  • 8 minutes reading time
  • Checked by StudySmarter Editorial Team
Save Article Save Article
Contents
Contents

Jump to a key chapter

    Importance of Limit Laws

    You may ask yourself why limit laws in calculus are important. You already know the definition of the limit of a function. Why not just apply that? The reason is that it is much more efficient to prove one thing about functions in general than to use the definition on each and every function. It is the difference between proving that dogs like to play and proving that my dog likes to play, your dog likes to play, the neighbor's dog likes to play... and on and on and on.

    Limit Laws in Calculus

    Many textbooks will mention the properties of limits listed below since they are the most common ones. Sometimes they will even refer to them as the 5 limit laws.

    Theorem: Properties of Limits, also known as Limit Laws

    Suppose that \(L\), \(M\), \(a\) and \(k\) are real numbers, with \(f\) and \(g\) being functions such that:

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

    and

    \[lim_{x \rightarrow a} g(x)=M\]

    Then the following hold:

    Sum Rule: \(lim_{x \rightarrow a} (f(x)+g(x))=L+M\)

    Difference Rule: \(lim_{x \rightarrow a} (f(x)-g(x))=L-M\)

    Product Rule: \(lim_{x \rightarrow a} (f(x)+ \cdot g(x))=L \cdot M\)

    Constant Multiple Rule: \(lim_{x \rightarrow a} k \cdot f(x) = k \cdot L\)

    Quotient Rule: If \neq 0\) then

    \[lim_{x \rightarrow a} \dfrac{f(x)}{g(x)}=\dfrac{L}{M}\]

    Power Rule: If \(r, s \in R \), with \(s \neq 0\), then

    \[lim_{x \rightarrow a} \left( f(x) \right)^{r/s}=L^{r/s} \]

    provided that \(L^{r/s}\) is a real number and \(L>0\) when \(s\) is even.

    For more examples of how to find limits of particular functions, see Finding Limits. For a reminder on the definition of the limit of a function, see Limits of a Function.

    It is essential to make sure that the conditions are met before applying the properties of limits. Let's see an example.

    Take

    \(f(x)=x\) and \(g(x)=\dfrac{1}{x}\) and try to find:

    \[lim_{x \rightarrow 0} (f \cdot g)(x)\].

    Answer:

    You are probably tempted just to use the Product Rule for limits. You already know that

    \[lim_{x \rightarrow 0} f(x)=0\]

    However, if you try and apply the definition of the limit for \(g(x)\), you can see that no matter how close you take your \(\delta\) window to be to \(a=0\), it won't work because the function has a vertical asymptote at \(x=0\). So, \(g(x)\) doesn't have a limit at \(a=0\). But

    \[lim_{x \rightarrow 0} (f(x) \cdot g(x))= lim_{x \rightarrow 0} \left( x \cdot \dfrac{1}{x} \right)\]

    \[lim_{x \rightarrow 0} (f(x) \cdot g(x))=lim_{x \rightarrow 0} (1)\]

    \[lim_{x \rightarrow 0} (f(x) \cdot g(x))=1\]

    which is not what you get when you multiply together \(0\) and something that doesn't exist! So while the limit of the product does exist, the product of the limits does not.

    Calculating Limits Using the Limit Laws

    For some functions, the limit laws get used so much that it is easier to look at kinds of functions rather than at lots of functions. It turns out that polynomials and rational functions are especially nice.

    Definitions and Limit Laws

    In the following examples, the definition of the limit was used to show that

    \[lim_{x \rightarrow a} x=a\]

    and

    \[lim_{x \rightarrow a} k=k\]

    where \(k\) is a constant. See Limits of a Function for more details on how to apply the definition of the limit.

    Take the function

    \[f(x)=10x^3-2x+1\]

    and \(a\) to be a constant real number. Find

    \[lim_{x \rightarrow a} f(x)\]

    Answer:

    Looking carefully, you can notice that the function is just the sum and product of powers of \(x\), along with the constant! So, the condition required to use our limit laws is met! Applying them gives:

    \[lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} (10x^3-2x+1)\]

    \[lim_{x \rightarrow a} f(x)= 10 (lim_{x \rightarrow a} x) (lim_{x \rightarrow a} x) (lim_{x \rightarrow a} x)- 2 (lim_{x \rightarrow a} x)+ (lim_{x \rightarrow a} 1) \]

    \[10a^3-2a+1\]

    In the example above, you looked at a specific polynomial and found the limit exists. It turns out that you can do this same process (using the Sum Rule, Constant Rule, and the Power Rule) to find the limit of any polynomial!

    If \(f(x)\) is a polynomial and \(a\) is a real number, then

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

    Limits of Rational Functions

    Taking the limit of rational functions can sometimes be a challenge. For more examples of techniques you can use for rational functions, see Finding Limits of Specific Functions.

    In the case where the point you want to take the limit at is in the domain of the rational function, taking the limit is not difficult. Take

    \[f(x)= \dfrac{x^2+3}{x-1}\]

    and find the limit as \(x \rightarrow 4\).

    Answer:

    First, ask yourself if \(x=4\) is in the domain of the function. It turns out that it is, and in fact

    \[f(4)=\dfrac{4^2+3}{4-1}=\dfrac{16+3}{3}=\dfrac{19}{3}\].

    So using the Quotient Rule tells you that

    \[lim_{x \rightarrow 4} = \dfrac{19}{3}\]

    Similar to the result for polynomials, you can say the following about rational functions:

    If \(f(x)\) is a rational function and \(a\) is a real number in the domain of \(f(x)\), then

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

    Examples Using the Limit Laws

    Rather than looking at a function with a definition, sometimes all you will know are some properties of the functions involved, and you will need to use Limit Laws to draw conclusions about the functions.

    Suppose that

    \(lim_{x \rightarrow 7} f(x)=3\) and \(lim_{x \rightarrow 7} g(x) =-1\)

    If possible, find the following:

    \[lim_{x \rightarrow 7} (f+g)(x)\]

    \[lim_{x \rightarrow 7} 4g(x)\]

    \[lim_{x \rightarrow 7} \left( \dfrac{f}{g} \right)(x)\]

    and

    \[lim_{x \rightarrow 7} \sqrt{g(x)}\]

    Answer:

    1. To find

    \[lim_{x \rightarrow 7} (f+g)(x)\]

    you have all the conditions satisfied to apply the Sum Rule, so

    \[lim_{x \rightarrow 7} (f+g)(x)=lim_{x \rightarrow 7} f(x) + lim_{x \rightarrow 7} g(x)\]

    \[lim_{x \rightarrow 7} (f+g)(x)=3+(-1)\]

    \[lim_{x \rightarrow 7} (f+g)(x)=2\]

    2. You can use the Constant Rule for the next one, so

    \[lim_{x \rightarrow 7} 4g(x)=4 lim_{x \rightarrow 7} g(x)\]

    \[lim_{x \rightarrow 7} 4g(x)=4(-1)\]

    \[lim_{x \rightarrow 7} 4g(x)=-4\]

    3. Since the limit of \(g(x)\) as \(x \rightarrow 7\) is not equal to zero, you can apply the Quotient Rule to see that

    \[lim_{x \rightarrow 7} \left( \dfrac{f}{g} \right)=\dfrac{\lim_{x \rightarrow 7}f(x)}{lim_{x \rightarrow 7} g(x)}=\dfrac{3}{-1}=-3\]

    4. The last one is a bit more challenging. Here,

    \[lim_{x \rightarrow 7} \sqrt{g(x)}=lim_{x \rightarrow 7}\sqrt{-1}\]

    You cannot take the square root of a negative number and get a real number back, so you can't evaluate this. That means you can't find

    \[lim_{x \rightarrow 7} \sqrt{g(x)}\]

    Limit Laws - Key takeaways

    • Suppose that \(L\), \(M\), \(a\) and \(k\) are real numbers, with \(f\) and \(g\) being functions such that:

      \(lim_{x \rightarrow a} f(x)=L\) and \(lim_{x \rightarrow a} g(x)=M\)

      Then the following hold:

      Sum Rule: \(lim_{x \rightarrow a} (f(x)+g(x))=L+M\)

      Difference Rule: \(lim_{x \rightarrow a} (f(x)-g(x))=L-M\)

      Product Rule: \(lim_{x \rightarrow a} (f(x) \cdot g(x))L \cdot M\)

      Constant Multiple Rule: \(lim_{x \rightarrow a} k \cdot f(x)=k \cdot L\)

      Quotient Rule: If \(M \neq 0\) then \[lim_{x \rightarrow a} \dfrac{f(x)}{g(x)}=\dfrac{L}{M}\]

      Power Rule: If \(r, s, \in Z\), with \(s \neq 0\), then \[lim_{x \rightarrow a} (f(x))^{r/s}=L{r/s}\]

      provided that \(L^{r/s}\) is a real number and \(L>0\) when \(s\) is even.

    • If \(f(x)\) is a polynomial and \(a\)is a real number, then \[lim_{x \rightarrow a} f(x)=f(a)\]
    • If \(f(x)\) is a rational function and \(a\) is a real number in the domain of \(f(x)\), then \[lim_{x \rightarrow a} f(x)=f(a)\]

    • Always be sure that the conditions to use one of the Limit Laws are met before you use it!

    Limit Laws Limit Laws
    Learn with 0 Limit Laws flashcards in the free StudySmarter app

    We have 14,000 flashcards about Dynamic Landscapes.

    Sign up with Email

    Already have an account? Log in

    Frequently Asked Questions about Limit Laws

    What are the 5 limit laws? 

    There are actually a lot more than 5 theorems about limits, but you probably mean the sum/difference rule, the constant multiple rule, the product rule, the quotient rule, and the power rule.

    How do you find a limit law? 

    Experimentation and research.

    How do you prove the sum law of limits? 

    By using the fact that both functions have a limit which is an actual number.

    What are limit laws? 

    They are really properties of limits that you can use so you don't have to show each and every different function has a limit.

    How many laws of limits are there? 

    Lots!  Most textbooks will talk about 5 main ones, but there are actually more of them.

    Save Article

    Discover learning materials with the free StudySmarter app

    Sign up for free
    1
    About StudySmarter

    StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

    Learn more
    StudySmarter Editorial Team

    Team Math Teachers

    • 8 minutes reading time
    • Checked by StudySmarter Editorial Team
    Save Explanation Save Explanation

    Study anywhere. Anytime.Across all devices.

    Sign-up for free

    Sign up to highlight and take notes. It’s 100% free.

    Join over 22 million students in learning with our StudySmarter App

    The first learning app that truly has everything you need to ace your exams in one place

    • Flashcards & Quizzes
    • AI Study Assistant
    • Study Planner
    • Mock-Exams
    • Smart Note-Taking
    Join over 22 million students in learning with our StudySmarter App
    Sign up with Email