# Calculus/Infinite Limits

 ← Finite Limits Calculus Continuity → Infinite Limits

## Informal Infinite Limits

Another kind of limit involves looking at what happens to ${\displaystyle f(x)}$ as ${\displaystyle x}$ gets very big. For example, consider the function ${\displaystyle f(x)={\frac {1}{x}}}$ . As ${\displaystyle x}$ gets very big, ${\displaystyle {\frac {1}{x}}}$ gets very small. In fact, ${\displaystyle {\frac {1}{x}}}$ gets closer and closer to 0 the bigger ${\displaystyle x}$ gets. Without limits it is very difficult to talk about this fact, because ${\displaystyle x}$ can keep getting bigger and bigger and ${\displaystyle {\frac {1}{x}}}$ never actually gets to 0; but the language of limits exists precisely to let us talk about the behavior of a function as it approaches something - without caring about the fact that it will never get there. In this case, however, we have the same problem as before: how big does ${\displaystyle x}$ have to be to be sure that ${\displaystyle f(x)}$ is really going towards 0?

In this case, we want to say that, however close we want ${\displaystyle f(x)}$ to get to 0, for ${\displaystyle x}$ big enough ${\displaystyle f(x)}$ is guaranteed to get that close. So we have yet another definition.

Definition: (Definition of a limit at infinity)

We call ${\displaystyle L}$ the limit of ${\displaystyle f(x)}$ as ${\displaystyle x}$ approaches infinity if ${\displaystyle f(x)}$ becomes arbitrarily close to ${\displaystyle L}$ whenever ${\displaystyle x}$ is sufficiently large.

When this holds we write

${\displaystyle \lim _{x\to \infty }f(x)=L}$

or

${\displaystyle f(x)\to L\quad {\mbox{as}}\quad x\to \infty }$

Similarly, we call ${\displaystyle L}$ the limit of ${\displaystyle f(x)}$ as ${\displaystyle x}$ approaches negative infinity if ${\displaystyle f(x)}$ becomes arbitrarily close to ${\displaystyle L}$ whenever ${\displaystyle x}$ is sufficiently negative.

When this holds we write

${\displaystyle \lim _{x\to -\infty }f(x)=L}$

or

${\displaystyle f(x)\to L\quad {\mbox{as}}\quad x\to -\infty }$

So, in this case, we write:

${\displaystyle \quad \lim _{x\to \infty }{\frac {1}{x}}=0}$

and say "The limit, as ${\displaystyle x}$ approaches infinity, equals ${\displaystyle 0}$ ," or "as ${\displaystyle x}$ approaches infinity, the function approaches 0.

We can also write:

${\displaystyle \lim _{x\to -\infty }{\frac {1}{x}}=0}$

because making ${\displaystyle x}$ very negative also forces ${\displaystyle {\frac {1}{x}}}$ to be close to ${\displaystyle 0}$ .

Notice, however, that infinity is not a number; it's just shorthand for saying "no matter how big." Thus, this is not the same as the regular limits we learned about in the last two chapters.

## Limits at Infinity of Rational Functions

One special case that comes up frequently is when we want to find the limit at ${\displaystyle \infty }$ (or ${\displaystyle -\infty }$) of a rational function. A rational function is just one made by dividing two polynomials by each other. For example, ${\displaystyle f(x)={\frac {x^{3}+x-6}{x^{2}-4x+3}}}$ is a rational function. Also, any polynomial is a rational function, since ${\displaystyle 1}$ is just a (very simple) polynomial, so we can write the function ${\displaystyle f(x)=x^{2}-3}$ as ${\displaystyle f(x)={\frac {x^{2}-3}{1}}}$ , the quotient of two polynomials.

Consider the numerator of a rational function as we allow the variable to grow very large (in either the positive or negative sense). The term with the highest exponent on the variable will dominate the numerator, and the other terms become more and more insignificant compared to the dominating term. The same applies to the denominator. In the limit, the other terms become negligible, and we only need to examine the dominating term in the numerator and denominator.

There is a simple rule for determining a limit of a rational function as the variable approaches infinity. Look for the term with the highest exponent on the variable in the numerator. Look for the same in the denominator. This rule is based on that information.

• If the exponent of the highest term in the numerator matches the exponent of the highest term in the denominator, the limit (at both ${\displaystyle \infty }$ and ${\displaystyle -\infty }$) is the ratio of the coefficients of the highest terms.
• If the numerator has the highest term, then the fraction is called "top-heavy". If, when you divide the numerator by the denominator the resulting exponent on the variable is even, then the limit (at both ${\displaystyle \infty }$ and ${\displaystyle -\infty }$) is ${\displaystyle \infty }$ . If it is odd, then the limit at ${\displaystyle \infty }$ is ${\displaystyle \infty }$ , and the limit at ${\displaystyle -\infty }$ is ${\displaystyle -\infty }$ .
• If the denominator has the highest term, then the fraction is called "bottom-heavy" and the limit at both ${\displaystyle \pm \infty }$ is 0.

Note that, if the numerator or denominator is a constant (including 1, as above), then this is the same as ${\displaystyle x^{0}}$ . Also, a straight power of ${\displaystyle x}$ , like ${\displaystyle x^{3}}$ , has coefficient 1, since it is the same as ${\displaystyle 1x^{3}}$ .

### Examples

Example 1

Find ${\displaystyle \lim _{x\to \infty }{\frac {x-5}{x-3}}}$ .

The function ${\displaystyle f(x)={\frac {x-5}{x-3}}}$ is the quotient of two polynomials, ${\displaystyle x-5}$ and ${\displaystyle x-3}$ . By our rule we look for the term with highest exponent in the numerator; it's ${\displaystyle x}$ . The term with highest exponent in the denominator is also ${\displaystyle x}$ . So, the limit is the ratio of their coefficients. Since ${\displaystyle x=1x}$ , both coefficients are 1, ${\displaystyle \lim _{x\to \infty }{\frac {x-5}{x-3}}={\frac {1}{1}}=1}$ .

Example 2

Find ${\displaystyle \lim _{x\to \infty }{\frac {x^{3}+x-6}{x^{2}-4x+3}}}$ . using lohospitals rule x^3+x-6= 3x^2+1 and x^2-4x+3 = 2x-4 =(3x^2+1)/(2x-4) = 1/-4

We look at the terms with the highest exponents; for the numerator, it is ${\displaystyle x^{3}}$ , while for the denominator it is ${\displaystyle x^{2}}$ . Since the exponent on the numerator is higher, we know the limit at ${\displaystyle \infty }$ will be ${\displaystyle \infty }$ . So,

${\displaystyle \lim _{x\to \infty }{\frac {x^{3}+x-6}{x^{2}-4x+3}}=\infty }$ .
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