\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - One-Sided Limits

Limits

Using Limits


Derivatives

Graphing

Related Rates

Optimization

Other Applications

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Calculus Tools

Learning Tools

Articles

One-sided limits require a good understanding of piecewise functions. If you are a little rusty or just need a quick reminder, you can find a complete discussion of piecewise functions on the piecewise functions precalculus page.

One-sided limits are finite limits where we evaluate the limit from each side of a point individually. The notation we use is

\(\displaystyle{ \lim_{x \to a^-}{f(x)} }\)

evaluate the limit on the left side of \(a\), i.e. values of \(x < a\)

\(\displaystyle{ \lim_{x \to a^+}{f(x)} }\)

evaluate the limit on the right side of \(a\), i.e. values of \(x > a\)

The negative and positive sign that look like exponents on the finite value \(a\) indicate the side that we are looking at.

One of the reasons we need to look at limits on both sides of some number is when we are determining continuity. As you know from the continuity page, one of the requirements for continuity is that the limit at a point must exist. In order for a limit to exist, the limit from the left must be equal to the limit from the right, i.e. \(\displaystyle{\lim_{x \to a^-}{f(x)} = \lim_{x \to a^+}{f(x)}}\).

Notice that we are NOT saying that the function value must be equal to the limit or even that the function need be defined at \(x=a\), only that the limit be equal on both sides of a.

Here is a great video to build your intuition of one-sided limits. He uses an absolute value function to discuss the idea of one-sided limits and limits that do not exist.

Dr Chris Tisdell - Limit of a function [9min-11secs]

video by Dr Chris Tisdell

Here is a good video showing a graph with several one-sided limits.

Krista King Math - How to find limits on CRAZY GRAPHS [7min-47secs]

video by Krista King Math

You will see more one-sided limits when you learn about continuity. For now, let's work these practice problems.

Practice

Unless otherwise instructed, evaluate these limits. Give your answers in exact form.

Basic

\(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)

Solution

484 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to0^-}{\left(\frac{1}{x}-\frac{1}{|x|}\right)}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0^-}{\left(\frac{1}{x}-\frac{1}{|x|}\right)}}\)

Solution

485 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Solution

486 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Solution

487 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)

Problem Statement

Evaluate the following limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)

Solution

489 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)

Solution

490 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)

Solution

491 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to0}{1/x}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{1/x}}\)

Solution

495 video

video by Khan Academy

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to0}{1/x^2}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{1/x^2}}\)

Solution

496 video

video by Khan Academy

close solution

Log in to rate this practice problem and to see it's current rating.

Intermediate

Evaluate \(\displaystyle{\lim_{x\to1}{f(x)}}\) for \(\displaystyle{f(x)=\left\{\begin{array}{lr} x+3 & x \leq 1 \\ x^2-2x & x >1 \end{array}\right.}\)

Problem Statement

Evaluate \(\displaystyle{\lim_{x\to1}{f(x)}}\) for \(\displaystyle{f(x)=\left\{\begin{array}{lr} x+3 & x \leq 1 \\ x^2-2x & x >1 \end{array}\right.}\)

Solution

488 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.

Problem Statement

Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.

Solution

492 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\)

Solution

494 video

video by Khan Academy

close solution

Log in to rate this practice problem and to see it's current rating.

one-sided limits 17calculus youtube playlist

You CAN Ace Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

learning and study techniques

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Shop eBags.com, the leading online retailer of luggage, handbags, backpacks, accessories, and more!

Try AmazonFresh Free Trial

As an Amazon Associate I earn from qualifying purchases.

How to Become a Straight-A Student: The Unconventional Strategies Real College Students Use to Score High While Studying Less

Shop eBags.com, the leading online retailer of luggage, handbags, backpacks, accessories, and more!

Shop Amazon - Sell Us Your Books - Get up to 80% Back

As an Amazon Associate I earn from qualifying purchases.

Practice Instructions

Unless otherwise instructed, evaluate these limits. Give your answers in exact form.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.