Що таке границі функцій на нескінченності?

Sometimes we want to say that a function gets close to some number when $x$ gets big. For example, consider the functions $f(x) = 2^{-x/3} + 2$ and $g(x) = x/2$, plotted below.

Unfortunately, the phrases "gets close to" and "flattens out" are vague. Instead, we use one of the below phrases to describe the behavior of the function.

$\bullet$ $f(x)$ approaches $2$ as $x$ approaches $\infty$.

$\bullet$ $f(x)\rightarrow 2$ as $x\rightarrow\infty$. (The arrow means "approaches".)

$\bullet$ The limit of $f(x)$, as $x$ approaches $\infty$, is $2$.

$\bullet$ $\displaystyle\lim_{x\rightarrow\infty}f(x) = 2$. ("lim" means "limit," and you read this equation exactly as the bullet above)

You can also just say

$\bullet$ $f(x)$ has a limit as $x$ approaches $\infty$.

without specifying what the limit is.

Consider the function over the reals $h(x) = \frac{1}{1 + x^2}$, plotted below.

Now we want to give a definition for the limit of a function over the reals.

Definition. If $f(x)$ is a function over the reals, then we write $\lim_{x\rightarrow\infty}f(x) = a$ if, for any $\epsilon > 0$, there exists a number $x_0$, such that $a - \epsilon < f(x) < a + \epsilon$ for all $x\ge x_0$. (We can also say $|a-f(x)|<\epsilon$).

Now look at the function $j(x)=f(-x)$.

But $j(x)$ gets closer and closer to $2$ as $x$ gets more negative. We say this in one of the following ways:

$\bullet$ $j(x)$ approaches $2$ as $x$ approaches $-\infty$.

$\bullet$ $j(x)\rightarrow 2$ as $x\rightarrow-\infty$.

$\bullet$ The limit of $j(x)$, as $x$ approaches $-\infty$, is $2$.

$\bullet$ $\displaystyle\lim_{x\rightarrow-\infty}j(x) = 2$.

Definition. $\lim_{x\rightarrow-\infty}f(x) = a$ if, for any $\epsilon > 0$, there exists a number $x_0$, such that $a - \epsilon < f(x) < a + \epsilon$ for all $x\le x_0$.

Hi. So we've studied the two-sided limits. Now ready for something new? (a.k.a one-sided limits)

One-sided limit sometimes means the limit when $x\rightarrow+\infty$ or $x\rightarrow-\infty$. This is intuitively true since you can't really approach positive infinity from "its right". You can only go from small values to big values of $x$, then "infinity".

Want an example?

So what would you expect when $x$ becomes very big? $y$ becomes really, really close to 0. So we can write that $lim_{x\to+\infty}\frac{1}{x}=0$.

You get it now. Here's another part of one-sided limits.

Here's a graph for reference to then next question:

So $y$ goes to two different values when $x$ approaches 0 from two different sides. Each one-sided limit can indeed be used separately if we only care about one side (like in geometry, everything needs to be positive.) But if we consider the function as a whole, the limit at $x=0$ can't take two values! So we usually say that there's NO LIMIT at this point.

In fact, two-sided limit exists only if both one-sided limit exists and agrees with each other.

This fact will become handy later, like a bent line (two lines glued together at a point) can not be differentiated at the point when one line changes into another.

(Video) Limits to Infinity

by A3Academy

Limits to Infinity - Introduction

First, watch this introduction on Limits to Infinity.

Now, answer some questions.

We can find the limit by diving the numerator and denominator with $$x^{32}$$ which makes all the terms with x in the denominator to zero,

$$\begin{align} \lim_{x \to \infty} \frac {x^{32}-3x^{15}+2}{3x^{32}-4x} = \lim_{x \to \infty} \frac {1- \frac {3}{x^{17}}+\frac {2}{x^{32}}} {3-\frac {4}{x^{31}}} = \frac {1} {3} \end{align}$$

As the x gets closer to 1 from below (because the limit is 1-, x-1 becomes more small and negative. Say substitute, 0.9, 0.99,0.999... So, you'd know whether the limit is finite or not.

$$\begin{align} \lim_{x \to 1-} \frac {1}{(x-1)} = -\infty \end{align}$$

You are very smart! Superb. Here is the last challenge for you. Let's see whether you can answer this.

Here x gets closer to -7 from below. Hence, $$\begin{align} \lim_{x\to -7^-} \frac {3} {x+7} = -\infty \end{align}$$

(Video) Limits at Infinity

by PatrickJMT

Here's a video by patrickJMT showing you how to find limits of functions as they approach infinity.

When we take a limit as it approaches infinity we're looking at what happens when we put in larger and larger numbers.

The first example he looks at is: $$\lim_{x\rightarrow\infty}\frac{3x^2+5x+4}{x^3+7x}$$ So we start putting in larger and larger numbers $$100,1000,1000000,1000000000,...$$ and if the output starts getting closer and closer to a number that's what we say the limit is.

When solving limits that approach infinity for rational functions, you always look at the highest power in the denominator. $$\frac{3x^2+5x+4}{\color{red}{x^3}+7x}$$ Here $x^3$ is our highest power in the denominator, so what we do is divide every term in our function by $x^3$. $$\begin{align} &=\lim_{x\rightarrow\infty}\frac{\frac{3x^2}{x^3}+\frac{5x}{x^3}+\frac{4}{x^3}}{\frac{x^3}{x^3}+\frac{7x}{x^3}}\\ &=\lim_{x\rightarrow\infty}\frac{\frac3x+\frac5{x^2}+\frac4{x^3}}{1+\frac7{x^2}} \end{align}$$ As $x$ gets closer to infinity, $\frac3x$ gets closer to zero. Same thing goes for all the terms with an $x$ in the denominator. $$=\lim_{x\rightarrow\infty}\frac{0+0+0}{1+0}=\frac01=0$$

So therefore, $$\lim_{x\rightarrow\infty}\frac{3x^2+5x+4}{x^3+7x}=0$$

The next example he solves is $$\lim_{x\rightarrow-\infty}\frac{x^4+x}{5x^3+7}$$ Here's the highest term in the denominator is $x^3$ so we divide each term by this. $$\begin{align} &=\lim_{x\rightarrow-\infty}\frac{\frac{x^4}{x^3}+\frac{x}{x^3}}{\frac{5x^3}{x^3}+\frac7{x^3}}\\ &=\lim_{x\rightarrow-\infty}\frac{x+\frac1{x^2}}{5+\frac7{x^3}} \end{align}$$ Like before, any term with an $x$ in the denominator becomes zero. $$=\lim_{x\rightarrow-\infty}\frac{-\infty+0}{5+0}=\frac{-\infty}5=-\infty$$ So therefore, $$\lim_{x\rightarrow-\infty}\frac{x^4+x}{5x^3+7}=-\infty$$

There are 3 different cases when dealing with the limits of rational functions as they approach infinity.

1. Degree in Numerator = Degree in Denominator. The limit is the ratio of the coefficients. $\\ \lim\limits_{x\rightarrow\pm\infty}=\dfrac{\color{red}{4}x^{10}+10000x^9}{\color{red}{5}x^{10}+4}=\dfrac45$
2. Degree in Numerator < Degree in Denominator. The limit is always zero. $\\ \lim\limits_{x\rightarrow\pm\infty}\dfrac{3x^7}{5x^8+10x+2}=0$
3. Degree in Numerator > Degree in Denominator. The limit will be either positive or negative infinity. $\\ \lim\limits_{x\rightarrow\infty}\dfrac{x^4}{x^3+5}=\infty$

(Video) Calculating a Limit at Infinity with a Radical

by PatrickJMT

Here's a video by patrickJMT showing you how to find the limit that approaches infinity of rational expressions when there's a radical in the denominator.

The example he looks at is $$\lim_{x\rightarrow\infty}\frac{x+3}{\sqrt{x^2+4}}$$ The approach is pretty much the same before. We pick out the largest power in the denominator. $$\frac{x+3}{\sqrt{\textcolor{red}{x^2}+4}}$$ Notice, though, the $x^2$ is underneath a radical. So if we take the square root of this we get: $$\sqrt{x^2}=x$$ So really, the highest power in the denominator is $x$. We want to divide every term by $x$. $$=\lim_{x\rightarrow\infty}\frac{\frac1x(x+3)}{\frac1x\sqrt{x^2+4}}$$ If we distribute in the numerator we get $$\frac{1+\frac3x}{}$$ In the denominator we have to be more careful. We have to convert $\frac1x$ into a radical in order to bring it into the other radical. $$\frac1x=\sqrt{\frac1{x^2}}$$ If we distribute this into the radical we get $$=\lim_{x\rightarrow\infty}\frac{1+\frac3x}{\sqrt{1+\frac4{x^2}}}$$ Plugging in infinity all the terms with $x$ in their denominator become zero. $$=\lim_{x\rightarrow\infty}\frac{1+0}{\sqrt{1+0}}=\frac1{\sqrt1}=1$$ So therefore, $$\lim_{x\rightarrow\infty}\frac{x+3}{\sqrt{x^2+4}}=1$$

The next example he looks at is $$\lim_{x\rightarrow-\infty}\frac{x^3}{\sqrt{x^6+4}}$$ Our highest power in the denominator is $x^6$ but since it's under the radical it becomes $x^3$. We can divide all terms by $x^3$. $$=\lim_{x\rightarrow-\infty}\frac{\frac{x^3}{x^3}}{\frac{1}{x^3}\sqrt{x^6+4}}$$ Just like in the last problem, we want to write $\frac1{x^3}$ as a radical. $$\frac1{x^3}=\sqrt{\frac1{x^6}}$$ But we have something different in this problem. Here, we have $x$ going to $-\infty$. So if we plug in negative numbers into $\dfrac1{x^3}$ we're going to get negative numbers back out. But if we put a negative number into $\sqrt{\dfrac1{x^6}}$ we're going to get a positive number out. So when we make this a radical we have to attach on a negative symbol in order to keep the number negative. $$=\lim_{x\rightarrow-\infty}\frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$ And now we continue on just like the problem above. $$=\lim_{x\rightarrow-\infty}\frac{1}{-\sqrt{1+\frac4{x^6}}}\\ =\lim_{x\rightarrow-\infty}\frac{1}{-\sqrt{1+0}}=\frac1{-\sqrt1}=-1$$ So, $$\lim_{x\rightarrow-\infty}\frac{x^3}{\sqrt{x^6+4}}=-1$$