Limits at Infinity

How to find limits at infinity?

In a section prior, we explored the concept of infinite limits. When we let x approach a number, sometimes we found that the value of f(x) became arbitrarily large (approaching infinity).

Now, we are going to let x become arbitrarily large (approach infinity) and investigate the behaviour of f(x).

Limits at infinity take on the following form:

Notation:

lim_{x \rightarrow \infty}f(x) = L

lim_{x \rightarrow -\infty}f(x) = L

To explore the concept of infinite limits, let’s look at the behaviour of the function:

\Large{f(x) = \frac{3x^2+1}{x^2+1}}

Similar to what we have done before, we will now let x approach a very large number, and see what happens to the values of f(x). The table shown summarizes this.

It is very clear that as x approaches a large number, the value of f(x) gets closer and closer to 3. So, we can say that:

lim_{x \rightarrow \infty}\frac{3x^2+1}{x^2+1} = 3

Then, it is clear that the function has a horizontal asymptote at y = 3. This leads us to the definition of the horizontal asymptote, as shown below:

Note:

The line y = L is the horizontal asymptote of f(x) if either:

lim_{x \rightarrow \infty}f(x) = L

lim_{x \rightarrow -\infty}f(x) = L

Example 8

Evaluate the following limit:

$\Large{\lim_{x \to \infty}(5x^3-2x^2-6x)}$

Solution

To solve this limit, our first idea would be to substitute $\infty$ into the function and see how it behaves. Well, let’s try that:

$\large{\lim_{x \to \infty}(5x^3-2x^2+6x) = 5(\infty)^3-2(\infty)^2+6(\infty)}$

\large{= \infty \:- \infty \:- \infty}

Be careful here. You might think the answer is zero or maybe $-\infty$ . However, this is not the correct answer and in fact, this is another one of those indeterminate forms. Hence, we need to solve this limit in a slightly different way.

First, we will factor out the highest degree of $x$ from all of the terms, as follows:

\large{\lim_{x \to \infty}(5x^3-2x^2-6x) = \lim_{x \to \infty}[x^3\:(5 \:- \frac{2}{x} \:- \frac{6}{x^2})]}

We can rewrite this as follows:

\large{\lim_{x\to \infty}(x^3) \:\cdot\:\lim_{x\to \infty}(5\:-\frac{2}{x}\:-\frac{6}{x^2})}

Finally, we have:

$\large{= \infty \:\cdot \: 5 = \infty}$ .

Thus,

\large{\lim_{x\to \infty}(5x^3-2x^2-6x) = \infty}