"একদিকবর্তী" সীমা

When we were looking at two-sided limits, we pointed out several examples that did not have limits to better understand what limits were. We promised, however, that one of those examples did have one-sided limits. Let's review that function quickly:

And once again, we look at the graph of the function to see what's going on.

No one value can "bridge the gap" between the positive part of this function and the negative part. That's why it doesn't have a limit at $x = 0$. Nevertheless, there are two conspicuous circles on the graph that suggest possible values for it at $x = 0$ that would at least make half of the graph nice. Surely those circles are meaningful in calculus.

Yes: they are exactly the one-sided limits.

Intuitively, a one-sided limit at $a$ is the value that would make one side of the function continuous --- either the side where $x \geq a$ or the side where $x \leq a$. Defining that value for a function at that point may not make the function fully continuous, but it's better than adding a point floating in the middle of nowhere.

Or, to use the idea of a limit as "getting really close", a one-sided limit is only different in that we restrict that $x$ must "get really close" from one side of the point.

The idea is that if $x$ gets close to $0$ but is still greater than $0$, then $f(x)$ gets really close to $1$. (In this example, it is $1$, but that's not necessary.) Note that we don't care about the values of $f(x)$ when $x = 0$ (just like for two-sided limits) or when $x < 0$.

If you understand this, the other side should be very easy:

We write the limit of $f(x)$ when $x$ approaches $a$ from above as $$\lim_{x \to a^+} f(x),$$ and when $x$ approaches $a$ from below, as $$\lim_{x \to a^-} f(x).$$ The important feature is the $+$ or $-$ superscript above $a$.

Let's look at more functions. This function should look familiar to you if you've done the questions on two-sided limits.

We see here a function that has one of its one-sided limits but not the other.

It seems it's about time to formalize the idea of a one-sided limit now. In fact, we only need to change a few words from the two-sided limit definition. And most of the changes are just to give the name of the concept we're defining (one-sided limits aren't quite as common as two-sided limits or limits at infinity in calculus, so a short name isn't very necessary.)

Reference definition: The limit of a function $f(x)$ at $a$ is a value $b$ such that, for all $\epsilon > 0$, we can find a number $\delta$ such that $f(x)$ is within $\epsilon$ of $b$ for all values of $x$ within $\delta$ of $a$ but not equal to $a$ itself.

Try changing the definition yourself before continuing:

Here are my definitions.

Definition: The limit of a function $f(x)$ at $a$ as $x$ approaches $a$ from above is a value $b$ such that, for all $\epsilon > 0$, we can find a number $\delta$ such that $f(x)$ is within $\epsilon$ of $b$ for all values of $x$ within $\delta$ of $a$ and greater than $a$ itself.

Definition: The limit of a function $f(x)$ at $a$ as $x$ approaches $a$ from below is a value $b$ such that, for all $\epsilon > 0$, we can find a number $\delta$ such that $f(x)$ is within $\epsilon$ of $b$ for all values of $x$ within $\delta$ of $a$ and less than $a$ itself.

(Equivalent ways to write the final condition on $x$ is that $a < x < a+\delta$ or $a-\delta < x < a$.)

So that's what a one-sided limit is.

A few final thoughts should be given to the relationship between two-sided limits and one-sided limits:

Congratulations! You now understand one-sided limits.

(Videos) One Sided Limits

by PatrickJMT

Here's a good video by patrickJMT showing you how to find one sided limits.

One sided limits are limits where you either approach a value from the left or the right of it. They are denoted with a minus or plus above the number being approached like so: $$\lim_{x\rightarrow a^-}\text{ or }\lim_{x\rightarrow a^+}$$

The first problem he looks at is: $$\text{Given }f(x)=\begin{cases}x+3;&x\le1\\x^2-2x;&x>1\end{cases}\\ \text{evaluate }\lim_{x\rightarrow1^-}f(x)$$ So notice our limit has a minus symbol above the $1$. This tells us we want to look at the values that approach 1 from the left. We can graph this to easily visualize.

So if we approach $1$ from the left we are taking numbers smaller than $1$. If we plug in these numbers we approach the $y$ value of $4$. $$\lim_{x\rightarrow1^-}f(x)=4$$

If you want to solve this algebraically, without graphing it, we just need to figure out which equation to use. Since we're approaching $1$ from the left we're taking numbers smaller than $1$. This means we use the equation $x+3$ because its condition is $x\le1$.

Since $x+3$ is continuous, we can just plug in $x=1$ into it to find the limit. $$\textcolor{red}{(1)}+3=4$$ As we can see this is the same answer as what we got when graphing.

If we were given $\lim\limits_{x\rightarrow1^+}f(x)$ instead, we would approach $1$ from the right. This means values larger than $1$. This would put us on the $x^2-2x$ piece. Since quadratics are continuous we can just plug in $x=1$ to find the limit. $$\textcolor{red}{(1)}^2-2\textcolor{red}{(1)}=-1$$

Finally, if the equation asked for $\lim\limits_{x\rightarrow1}f(x)$ where it didn't specify which side then we would say the limit does not exist because the limits need to approach the same number which they do not in this example.

In this video he looks at the next example: $$\text{Given }g(x)=|x+4|,\\\text{find}\lim_{x\rightarrow-4^-}g(x)$$

If we quickly graph this we get:

As we approach $-4$ from the left we're taking values smaller than $-4$. We see the limit begins to approach $0$.

Algebraically, we can just plug in $x=-4$. This is because the absolute value function is continuous. $$|\textcolor{red}{(-4)}+4|=0$$

If we approach $-4$ from the right, $\lim\limits_{x\rightarrow-4^+}g(x)$, we also get $0$ as an answer. Since the left hand limit equals the right hand limit we can say the limit at $-4$ is $0$. $$\lim_{x\rightarrow-4}g(x)=0$$

In this last video he looks at one final problem. $$\text{Given }h(x)=\sqrt{x^2-25},\\\text{find}\lim_{x\rightarrow5^+}h(x)$$

If we begin approaching $5$ from the right we take points a bit larger than five. If we put these into $\sqrt{x^2-25}$ and get closer and closer to $5$ we see the value starts getting closer to $0$. $$\sqrt{\color{red}{(5)}^2-25}=0$$

Be careful though, if we were asked to find $\lim\limits_{x\rightarrow5^-}h(x)$ the limit does not exist. This is because if we take values smaller than $5$ then we get negative values inside our square root.