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17calculus > limits > one-sided limits

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 17calculus 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

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]

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Practice Problems

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

Level A - Basic

Practice A01

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

solution

Practice A02

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

solution

Practice A03

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

solution

Practice A04

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

solution

Practice A05

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

solution

Practice A06

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

solution

Practice A07

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

solution

Practice A08

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

solution

Practice A09

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

solution


Level B - Intermediate

Practice B01

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

Practice B02

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

solution

Practice B03

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

solution

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