On this screen we consider one-sided limits.

Revisiting the function that “jumps” from the preceding screen
Recall one of the functions we considered on the previous screen for which the limit does not exist:
\[f(x) =
\begin{cases}
-3 & \text{for } x \lt 0 \\
\phantom{-}5 & \text{for } x \ge 0
\end{cases}\]
As you’ll remember, $\displaystyle{\lim_{x \to 0} f(x) }$ does not exist because the function “jumps” at $x=0$ from $-5$ to 5, and so there is no single number L that we can be as close to as we’d like for values of x both less than and greater than $x=0.$ That is, there is no single place on the graph you can point to that represents the limit at $x=0.$

You probably have the thought, though, that if we could consider only values of $x \lt 0,$ and remain always on the left-side of the vertical axis, then the limit does exist, and equals $-3$. If so, you’re right! Similarly, if we consider values of $x \gt 0$ and stay to the right of the vertical axis, then the limit again exists, and equals 5.

“Limit from the left” and “limit from the right”

By examining the behavior of a function on either one side or the other of a point of interest, we are considering one-sided limits. The only real thing to remember is what the notation means:

 
So for the function $f(x)$ above:

That is, we can get as close to the function’s output value of $-3$ as we’d like
by being sufficiently close to $x=0$ from the left.

And

That is, we can get as close to the function’s output value of 5 as we’d like
by being sufficiently close to $x=0$ from the right.

The general limit and one-sided limits at $x=a$

Still thinking about the function above, we can now write (very mathily!) using the one-sided limit notation:
\[\text{Because } \lim_{x \to 0^-}f(x) \ne \lim_{x \to 0+}f(x) \text{, } \lim_{x \to 0}f(x) = \text{DNE}\] In words: because the limit from the left does not equal the limit from the right, the (general) limit does not exist at $x=0.$ (We include the word “general” there in parentheses only to indicate that we do not mean either one-sided limit. Typically you wouldn’t write that word; instead the word “limit” alone means you have to consider what happens as you approach the point of interest from both directions.)

Indeed, it is always true that the (general) limit exists and equals L only if the limit from both sides equals the same value L:

 
Continuing to think about our example piecewise function above, $f(x),$ if we consider any point other than $x=0,$ then the (general) limit exists because the one-sided limits are equal.

For instance, at $x=-2,$
\[\lim_{x \to -2^-}f(x) = -3 \quad \text{and} \quad \lim_{x \to -2+}f(x)= -3 \] \[\text{and hence } \lim_{x \to -2}f(x) = -3.\] And at $x= 4,$
\[\lim_{x \to 4^-}f(x) =5 \quad \text{and} \quad \lim_{x \to 4^+}f(x)= 5 \] \[\text{and hence } \lim_{x \to 4}f(x) = 5.\]

Revisiting $g(x) = \dfrac{\sin(x)}{x}$
As a different example, let’s consider another function we looked at earlier,
\[g(x) = \dfrac{\sin(x)}{x}\]
For this function we have
\[\lim_{x \to 0^-}g(x) = 1 \quad \text{and} \quad \lim_{x \to 0^+}g(x) = 1 \] \[\text{and hence } \lim_{x \to 0}g(x) = 1.\] This is of course the same conclusion we reached earlier; we’re merely using the function to illustrate that when the one-sided limits are the same, the (general) limit exists, and equals the value of both one-sided limits.

Nothing more to discuss here; practice problems for you to try are below!

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Questions or comments about anything on this screen? Please let us know on the Forum!

The Upshot

On the next screen, we’ll consider some functions that grow without bound, heading to $\infty$ or $-\infty$ at a particular value of x.

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Practice

Question 1 is typical of homework and AP^® multiple-choice type questions.