Infinity:
Before talking about the limits at infinity, we should clear our concept about infinity. So, what is infinity?
Everyone defines by their own way but in the simplest words “ the infinity is something like endless which have not a certain closing or ending or boundary point, where it may end”.
Now think and observe it’s all above the natural or real things or numbers.
E.g think about stars in space can you count them?
Similarly, what about the hairs on your head?
Now you can understand that what actually is the limits. So, let's talk about limits at infinity.
Limits at Infinity:
We have an amazing example of limits approaching infinity,
What about? $$\frac{1}{stars}\;or\;\frac{1}{hairs}$$
It is the same as:
$$\frac{1}{\infty}$$
We have no knowledge of what actually it is?
Maybe it’s a 0 or undefined?
So, let’s try to know about the limit as x approaches infinity.
Finding Limits At Infinity:
Consider an example of limit as x approaches infinity:
$$\lim\limits_{x \to \infty} \frac{1}{x}$$
Let’s start plugging in a large and larger number, as it becomes countless or limits approaching infinity.
Whenever the denominator is large you're gonna get a small value if the denominator has a small value you get a large value.
| x | $$\frac{1}{x}$$ |
|---|---|
| 100 | 0.01 |
| 1,00,000 | 0.00001 |
| 10,000,000 | 0.0000001 |
| 10,000,000,000 | 0.0000000001 |
So, what we observed?
As the x approaches to infinity then 1/x become smaller and smaller which is near equals to 0.
In this way, we can say limits at infinity should be 0.
$$\lim\limits_{x \to \infty} \frac{1}{x}=0$$
As infinity is a very large number, it would take place the expression going to be 0.
Now, what about the limits approaching infinity, if we follow the same technique from the right-hand limit by using negative infinity?
Let’s see is it also 0 or function does not exist in such a way:
For the right-hand limit,
$$\lim\limits_{x \to \infty} \frac{1}{x}$$
| x | $$\frac{1}{x}$$ |
|---|---|
| -100 | -0.01 |
| -1,00,000 | -0.00001 |
| -10,000,000 | -0.0000001 |
| -10,000,000,000 | -0.0000000001 |
Now we can see that it still approaching 0 but from a negative end.
The result we conducted here
For left-hand limit
$$\lim\limits_{x \to +\infty} \frac{1}{x}=0^+$$
For right-hand limit
$$\lim\limits_{x \to -\infty} \frac{1}{x}=0^-$$
Graphical representation:
Now graph the function 1/x with limit as x approaches infinity, it looks like
The function 1/x havs vertical and horizontal asymptotes at x = 0 and at y = 0.
Horizontal Asymptotes : As the the indicates that horizontal asymptote is the horizontal line present in the graph of the function which approaches from +∞ to -∞ or boundary less.
Similarly,
Vertical Asymptotes are those which presents vertical boundless line in the graph of the function.
Find Horizontal Asymptote
So as you can see as x approaches positive infinity you're going to get the horizontal asymptote of a function which is y = 0. As we follow the curve all the way to the right it's going to get closer and closer to 0. This will also give you the horizontal asymptote on the left side which is also y=0.
$$\lim\limits_{x \to +\infty} \frac{1}{x}=0^+\;\text{Horizontal aysmptote}$$
Find Vertical Asymptote
To get the vertical asymptote as X approaches 0 from the right side. You're going to get negative infinity it's going down so the x value that leads to our y value of infinity that's going to be the vertical asymptote.
$$\lim\limits_{x \to -\infty} \frac{1}{x}=0^-\;\text{Vertical aysmptote}$$
When limit as x approaches infinity of function, then value of Y gives a graph which is the horizontal asymptote of a function.
Let's try more problem of vertical and horizontal asymptotes:
Example
what is the limit as x approaches infinity of :
$$\lim\limits_{x \to \infty} \frac{1}{x^2}$$
We have another bottom-heavy function and if we plug in large numbers
| x | $$\frac{1}{x^2}$$ |
|---|---|
| 100 | 0.0001 |
| 1000 | 0.000001 |
| 1,00,000 | 0.0000000001 |
| 10,00,000 | 0.000000000001 |
We could say this is going to approach a zero anytime you have 1/infinity. It's always going to equal zero.
General Theorem
So, the general theorem for finding limits at infinity is:
$$\lim\limits_{x \to \infty} \frac{1}{x^R}=0$$
Example
Let’s see an example
$$\lim\limits_{x \to \infty} \frac{8}{3x+4}$$
Divide numerator and denominator by x for 1/x form
$$\lim\limits_{x \to \infty} \frac{8}{x}\;÷\;\frac{3x+4}{x}$$ $$\Rightarrow \; \lim\limits_{x \to \infty} \frac{8}{x}\;÷\;(3+\frac{4}{x})$$ $$\Rightarrow \; \{ \lim\limits_{x \to \infty} \frac{8}{x} \} \; ÷ \; \{ \lim\limits_{x \to \infty}\;(3+\frac{4}{x}) \}$$
Since
$$\lim\limits_{x \to \infty} \frac{1}{x^R}=0$$
So,
$$ \lim\limits_{x \to \infty} \frac{8}{x}=0 \; \text{&} \; \lim\limits_{x \to \infty}(3+\frac{4}{x})=3+0 $$ $$\Rightarrow \; \{ \lim\limits_{x \to \infty} \frac{8}{x} \} \; ÷ \; \{ \lim\limits_{x \to \infty}\;(3+\frac{4}{x}) \} \;=\; \frac{0}{0+3}$$ $$\lim\limits_{x \to \infty} \frac{8}{3x+4} \;=\; \frac{0}{0} \;=\; 0$$
Yes, it is heading towards 0.