17Calculus - One-Sided Limits
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.
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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 |
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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 |
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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 |
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\(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)
Solution |
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484 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to0^-}{\left(\frac{1}{x}-\frac{1}{|x|}\right)}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0^-}{\left(\frac{1}{x}-\frac{1}{|x|}\right)}}\)
Solution |
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485 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)
Solution |
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486 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)
Solution |
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487 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)
Problem Statement |
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Evaluate the following limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)
Solution |
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489 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)
Solution |
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490 video
video by PatrickJMT |
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\(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)
Solution |
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491 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{1/x}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{1/x}}\)
Solution |
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495 video
video by Khan Academy |
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\(\displaystyle{\lim_{x\to0}{1/x^2}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{1/x^2}}\)
Solution |
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496 video
video by Khan Academy |
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Intermediate |
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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 |
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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 |
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488 video
video by PatrickJMT |
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Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.
Problem Statement |
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Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.
Solution |
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492 video
video by Krista King Math |
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\(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\)
Problem Statement |
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Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\)
Solution |
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494 video
video by Khan Academy |
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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
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, evaluate these limits. Give your answers in exact form.
