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one-sided limits youtube playlist

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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

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

One-Sided Limits - Practice Problems Conversion

[A01-484] - [A02-485] - [A03-486] - [A04-487] - [A05-489] - [A06-490] - [A07-491] - [A08-495] - [A09-496]

[B01-488] - [B02-492] - [B03-494]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, evaluate the following limits. Give your answers in exact form.

Basic Problems

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

Problem Statement

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

Solution

484 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

485 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

486 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

487 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

489 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

490 video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

491 video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

495 video

video by Khan Academy

close solution

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

Problem Statement

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

Solution

496 video

video by Khan Academy

close solution

Intermediate Problems

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

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

Problem Statement

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

Solution

492 video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

494 video

video by Khan Academy

close solution
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