Let $f(x)$ be some function (defined over the reals). How do I find the limit as $x$ approaches $a$ from either the positive or the negative side? (i.e how do I find either $\lim\limits_{x \rightarrow a^+} f(x)$ or$\lim\limits_{x \rightarrow a^- } f(x)$)


A one sided limit of $f(x)$ as $x$ approaches $a^+$ (a from the positive side, meaning on the interval $[a,\infty)$) or $a^-$ (a from the negative side, meaning on the interval $(- \infty, a]$) can be easily defined as the value which a function approaches as values of $x$ from the respective (positive or negative) side get progressively closer to $a$.

There are several techniques one can use to find a one sided limit:


To find a one sided limit graphically all that would be necessary is to visually see what the value of $f(x)$ is as $x$ gets arbitrarily close to $a$ from the positive side of the $x$-axis (for $\lim\limits_{x \rightarrow a^+} f(x)$) or from the negative side of the $x$-axis (for $\lim\limits_{x \rightarrow a^- } f(x)$). Take the following graph for example:

graph of y=x

To find the limit of $f(x)$ as $x \rightarrow 3^+$, you would start at some $x$ greater than $3$, and then make that $x$ closer and closer to $3$ observing any pattern in the value of $f(x)$, and then looking to see what the value $f(3)$ would be if $f(x)$ continued the same pattern as you have been observing. In this case I might start at four and notice that as I get closer and closer to $3$, $f(x)$ also gets closer to $3$, and I might therefore conclude that $$\lim\limits_{x \rightarrow a^+} f(x)= 3 \ \ .$$

Note: that if the graph has a hole at the point you are approaching, this will not change the limit. Similarly, if the graph gets arbitrarily bigger as you approach $a$, the limit is $\infty$, or if the graph gets arbitrarily more negative the limit is $- \infty$.


While this is not a general rule, if the function is "well behaved" (continuous), meaning it has not holes or jumps or anything like that, then simply plugging in $a$ will yield the limit. However, that is not a general way of finding the limit. A more general way would be to plug in values from whichever side the limit is approaching and see what value, if any $f(x)$ approaches.

An example of this might be $\lim \limits_{x \rightarrow 2^-} \frac{x}{x+2}$. While we could plug in $2$, because this is a "well behaved" function around $2$, if we plug in values closer and closer to $2$ from the negative side we can create a value table as follows (top is $x$, bottom is $f(x)$):

value table .

This shows us that the value of $f(x)$gets closer and closer to $\frac{1}{2}$ as $x$ gets closer to $2$.

Note: The same disclaimers as in the previous method apply here.


A more rigorous way of approaching this problem is through the delta-epsilon definition of the one sided limit, and although this is rarely the approach used in introductory calculus classes, it will be used in most proof-based calculus classes.

The definition of a one sided limit is (as taken from this question):

$\displaystyle \lim_{x \to a^+} f(x) = L$ if and only if For any $\epsilon>0$ there is a $\delta>0$ so that for any $x$, if $0 < x-a <\delta$ then $|f(x)-L| < \epsilon$.

$\displaystyle \lim_{x \to a^-} f(x) = L$ if and only if For any $\epsilon>0$ there is a $\delta>0$ so that for any $x$, if $0 < a-x <\delta$ then $|f(x)-L| < \epsilon$.

Using this you can prove that a certain $L$ is the limit.

Intuitively, this definition simply says that no matter how close you get to $L$, there is always a range of $x$ values which contain $a$ and which, when plugged in to $f(x)$ will be equally as close. In essence, no matter how close $f(x)$ gets to $L$, you can still "approach" it by making $x$ closer and closer to $a$ (from whichever side the limit dictates).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.