One-sided limits and the non-existence of a limit

We already know what a limit of a function is. Limits allow us to determine what value a function is approaching when we use a particular input. But what if the input $x$ approaches some number $a$ only on the right or left side? Could the limit not exist at all? In this topic, we will answer these questions.

One-sided limits

The limit of a function can depend on the direction in which the function is moving. Let's consider an example. Suppose

$$f(x) = \begin{cases} -1 \quad& x< 0 \\ 1 \quad& x \geqslant 0.\end{cases}$$What is $\lim\limits_{x \to 0} f(x)$ equal to? We get a different value if we get to $0$ from the positive or negative side:

Hence, there is no limit when $x \to 0$. But we can find two different one-sided limits. The limit of $f(x)$ as $x$ approaches 0 from the negative side, which is denoted by $\lim\limits_{x \to 0 -} f(x)$, is equal to $-1$:

$$\lim\limits_{x \to 0 -} f(x)=-1$$ The limit of $f(x)$ as $x$ approaches 0 from the positive side, which is denoted by $\lim\limits_{x \to 0 +} f(x)$, is equal to $1$:

$$\lim\limits_{x \to 0 +} f(x)=1$$These limits are called the left-hand and right-hand limits.

Non-existence of a limit

Note that there are cases when a limit does not exist. We will prove it with the definition of a limit according to Heine (we mentioned it in the topic "A limit of a function"). It is formulated this way:

$A$ is a limit of a function $f(x)$ in $x_0$ if for any sequence $x_n$ which is different from $x_0$ but tends to it, then the sequence of $f(x_n)$approaches $A$.

Let's take a look at this limit:

$$\lim\limits_{x \to 0} \sin \left(\dfrac{1}{x}\right)$$If a limit exists, then no matter how we approach $0$, the value of the function will approach a specific value.

Let's approach $0$ the following way: we consider the sequence

$$\dfrac{1}{2\pi n} \ \ n = 1, 2, 3, \ldots$$If $n \to \infty$, this sequence approaches $0$.

Substituting these values into the function, we obtain

$$\sin \left( \dfrac{1}{\frac{1}{2\pi n}} \right)= \sin (2\pi n) =0$$Now let's approach $0$ in another way: consider the sequence

$$\sin \left(\dfrac{1}{\dfrac{1}{\dfrac{\pi}{2} +2\pi n}}\right) = \sin \left( \dfrac{\pi}{2} + 2 \pi n \right)=1.$$We have shown that depending on how we approach $0$, the function $\sin \dfrac{1}{x}$ takes different values: $0$ and $1$. Hence, the limit $\lim\limits_{x \to 0} \sin \dfrac{1}{x}$ does not exist.

Infinite limits

Sometimes the function does not tend to a certain value and just increases or decreases infinitely. For example, $\dfrac{1}{x}$ tends to $\infty$ as we mentioned earlier. The infinite limits do not exist according to the previous definitions, but they still can help us in understanding the behavior of the function.

The definition of $+\infty$ limit is:

For any $N>0$ exists $\delta$ such that if $0<|x-a|<\delta$ then $f(x)>N$

For $-\infty$ it will be:

For any $N<0$ exists $\delta$ such that if $0<|x-a|<\delta$ then $f(x)<N$

And for $\infty$:

For any $N>0$ exists $\delta$ such that if $0<|x-a|<\delta$ then $|f(x)|>N$

In our example if $x$ approaches $0$ from the right side, then the function increases infinitely. While if $x$ approaches from the left side, then the function decreases infinitely. To sum it up we can say that function approaches infinity no matter how $x$ approaches $0$.

$\lim\limits_{x \to 0+} \dfrac{1}{x}=+\infty, right-hand\ \ limit\\ \lim\limits_{x \to 0-} \dfrac{1}{x}=-\infty, left-hand \ limit\\\lim\limits_{x \to 0} \dfrac{1}{x}=\infty$

Conclusion

In this topic, we learned about the left-hand and right-hand limits. These are the limits of a function, implying "approaching" the limit point from one side. We also looked at cases where a function limit may not exist and solved a few examples.