# How to solve a one sided limit?

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:

### Graphically:

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:

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 $$- \infty$$.

### Algebraically

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 $$f(x)$$):

.

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.

### Analytical

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).