Limits at Infinity

Here is the graph of $f(x) =
   \dfrac{x^2}{x^2 + 1}$ :

$$\hbox{\epsfysize=1.75in \epsffile{inflim1.eps}}$$

The graph approaches the horizontal line $y = 1$ as it goes out to the left and right. You write:

$$\lim_{x \to +\infty} \dfrac{x^2}{x^2 + 1} = 1 \quad\hbox{and}\quad \lim_{x \to -\infty} \dfrac{x^2}{x^2 + 1} = 1.$$

In general, to say that

$$\lim_{x \to +\infty} f(x) = L$$

means that the graph of $f(x)$ approaches $y =
   L$ as you plug in larger and larger positive values for x.

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

means that the graph of $f(x)$ approaches $y
   = L$ as you plug in larger and larger negative values for x.

For example, consider $f(x) =
   \dfrac{x^2}{x^2 + 1}$ . If you set $x = 10^6$ , you get

$$f(x) \approx 0.999999999999000000000000999999999999000000000001.$$

That's pretty close to 1, isn't it?

Let's look some examples of a limits at infinity.


Example. $\displaystyle \lim_{x \to +\infty} \dfrac{2x^3 - 3x + 7}{4 -
   x^2 - 5x^3}$

In limits at infinity involving powers of x, the rule of thumb is that the biggest powers dominate. The limit above behaves almost like

$$\lim_{x \to +\infty} \dfrac{2x^3}{-5x^3},$$

because the $x^3$ 's on the top and bottom dominate. you expect the answer to be $-\dfrac{2}{5}$ .

On way to see this formally is to divide the top and bottom by $x^3$ :

$$\lim_{x \to +\infty} \dfrac{2x^3 - 3x + 7}{4 - x^2 - 5x^3} = \lim_{x \to +\infty} \dfrac{2 - \dfrac{3}{x^2} + \dfrac{7}{x^3}} {\dfrac{4}{x^3} - \dfrac{1}{x} - 5}.$$

Now as $x \to +\infty$ ,

$$\dfrac{\hbox{a number}}{x^{\hbox{positive power}}} \to \dfrac{\hbox{a number}}{\hbox{something humongous}} = 0.$$

Hence,

$$\lim_{x \to +\infty} \dfrac{2 - \dfrac{3}{x^2} + \dfrac{7}{x^3}} {\dfrac{4}{x^3} - \dfrac{1}{x} - 5} = \lim_{x \to +\infty} \dfrac{2 - 0 + 0}{0 - 0 - 5} = -\dfrac{2}{5}.$$

Here's a picture of $\dfrac{2x^3 - 3x +
   7}{4 - x^2 - 5x^3}$ :

$$\hbox{\epsfysize=1.75in \epsffile{inflim2.eps}}$$

What else can happen?

$$\lim_{x \to -\infty} \dfrac{x^{15} - 3x^9 + 47}{x^2 - x + 1} = -\infty,$$

because the $x^{15}$ on top beats out the puny $x^2$ on the bottom.

By the way, it would be correct to say this limit diverges. However, it's more informative to say how it diverges. In this case, the function $\dfrac{x^{15} -
   3x^9 + 47}{x^2 - x + 1}$ becomes large and negative, so you write $-\infty$ for the limit.

On the other hand,

$$\lim_{x \to +\infty} \dfrac{x - 17}{x^{3/2} - 4x + 2} = 0,$$

because the $x^{3/2}$ on the bottom beats out the $x^1$ on the top.


I noted above that

$$\lim_{x \to +\infty} f(x) = L$$

means that the graph of $f(x)$ approaches the line $y = L$ as you move to the right, and

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

means that the graph of $f(x)$ approaches the line $y = L$ as you move to the left. In these situations, $y = L$ is a horizontal asymptote for the graph of $f(x)$ .

Not all graphs have horizontal asymptotes --- for example, $y = x^2$ goes to $\infty$ as $x\to \infty$ and as $x\to -\infty$ . You can check for the presence of horizontal asymptotes by computing $\lim_{x \to +\infty} f(x)$ and $\lim_{x
   \to -\infty} f(x)$ and seeing if either is a number.


Example. Find the horizontal asymptotes (if any) of $y = \dfrac{x}{x^2 + 1}$ .

Since

$$\lim_{x\to \infty} \dfrac{x}{x^2 + 1} = 0 \quad\hbox{and}\quad \lim_{x\to -\infty} \dfrac{x}{x^2 + 1} = 0,$$

$y = 0$ is a horizontal asymptote for the graph at $+\infty$ and at $-\infty$ . The graph is shown below:

$$\hbox{\epsfysize=1.75in \epsffile{inflim3.eps}}\quad\halmos$$


Example. Find the horizontal asymptotes of $f(x) = \dfrac{x}{\sqrt{x^2 + 4}}$ .

The limit at $+\infty$ works without any surprises. The highest power on the top and the bottom is x (since $\sqrt{x^2}$ looks like x), so divide the top and bottom by x:

$$\lim_{x \to +\infty} \dfrac{x}{\sqrt{x^2 + 4}} = \lim_{x \to +\infty} \dfrac{1}{\dfrac{1}{x}\sqrt{x^2 + 4}} = \lim_{x \to +\infty} \dfrac{1}{\sqrt{1 + \dfrac{4}{x^2}}} = \dfrac{1}{1} = 1.$$

However, the limit at $-\infty$ is a little tricky! Here's the computation:

$$\lim_{x \to -\infty} \dfrac{x}{\sqrt{x^2 + 4}} = \lim_{x \to -\infty} \dfrac{1}{\dfrac{1}{x}\sqrt{x^2 + 4}} = \lim_{x \to -\infty} \dfrac{1}{-\sqrt{1 + \dfrac{4}{x^2}}} = \dfrac{1}{-1} = -1.$$

Where did that negative sign come from? Look at the bottom, which was $\dfrac{1}{x}\sqrt{x^2 + 4}$ . x is going to $-\infty$ , so x is taking on negative values. Now $\sqrt{\cdot}$ is positive, so $\dfrac{1}{x}\sqrt{x^2 + 4}$ is negative.

When you push the $\dfrac{1}{x}$ into the square root, you must leave a negative sign outside. Otherwise, you'd have $\sqrt{\hbox{junk}}$ , a positive thing.

This is a case where it matters that x is going to $-\infty$ , as opposed to $+\infty$ . Here's the graph:

$$\hbox{\epsfysize=1.75in \epsffile{inflim4.eps}}\quad\halmos$$


How do logarithms and exponentials behave as $x\to +\infty$ or $x\to -\infty$ ? The relevant facts are summarized below.

$$\lim_{x\to +\infty} \ln ax = +\infty \quad\hbox{and}\quad \lim_{x\to 0^+} \ln ax = -\infty \quad\hbox{if}\quad a > 0.$$

$$\lim_{x\to +\infty} e^{ax} = +\infty \quad\hbox{and}\quad \lim_{x\to -\infty} e^{ax} = 0 \quad\hbox{if}\quad a > 0.$$

I've graphed $y = \ln 2x$ (on the left) and $y = e^{3x}$ (on the right) below; you can see that the pictures are consistent with the formulas above.

$$\hbox{\epsfysize=1.75in \epsffile{inflim5.eps}}\hskip0.5in \hbox{\epsfysize=1.75in \epsffile{inflim6.eps}}$$

For example, the graph of $y = \ln 2x$ goes downward asymptotically along the y-axis from the right. This confirms that $\displaystyle \lim_{x\to 0^+} \ln 2x = -\infty$ .

Likewise, the graph of $e^{3x}$ rises sharply as you go to the right; this confirms that $\displaystyle
   \lim_{x\to +\infty} e^{3x} = +\infty$ .

Note that if $a < 0$ in $e^{ax}$ , the limits are reversed. Specifically,

$$\lim_{x\to +\infty} e^{ax} = 0 \quad\hbox{and}\quad \lim_{x\to -\infty} e^{ax} = +\infty \quad\hbox{if}\quad a < 0.$$


Example.

$$\lim_{x\to +\infty} \ln 1.37x = +\infty \quad\hbox{and}\quad \lim_{x\to 0^+} \ln 1.37x = -\infty.$$

$$\lim_{x\to +\infty} e^{6x} = +\infty \quad\hbox{and}\quad \lim_{x\to -\infty} e^{6x} = 0.$$

$$\lim_{x\to +\infty} e^{-\sqrt{2}x} = 0 \quad\hbox{and}\quad \lim_{x\to -\infty} e^{-\sqrt{2}x} = +\infty.\quad\halmos$$


Infinity can also appear in limits in connection with vertical asymptotes. I'll say that the graph of a function $y = f(x)$ has a vertical asymptote at $x = a$ if at least one of the limits

$$\lim_{x\to a^+} f(x) \quad\hbox{or}\quad \lim_{x\to a^-} f(x)$$

is either $+\infty$ or $-\infty$ .


Example. The graph below has a vertical asymptote at $x = a$ :

$$\hbox{\epsfysize=1.75in \epsffile{inflim7.eps}}$$

In this case,

$$\lim_{x\to a^+} f(x) = -\infty \quad\hbox{while}\quad \lim_{x\to a^-} f(x) = +\infty.\quad\halmos$$


In general, you might suspect the presence of a vertical asymptote at an isolated value of x for which $f(x)$ is undefined. To confirm your suspicion, you need to compute the left- and right-hand limits at the point.


Example. Locate the vertical asymptotes of $f(x) = \dfrac{1}{(x - 1)(x - 2)}$ and sketch the graph near the asymptotes.

$f(x)$ is undefined at $x = 1$ and at $x = 2$ . I'll check for vertical asymptotes by computing the left- and right-hand limits at $x = 1$ and at $x = 2$ . I'll work through the first one carefully.

$$\lim_{x\to 1^+} \dfrac{1}{(x - 1)(x - 2)} = -\infty.$$

To see this, consider numbers close to 1 but to the right of 1. Then $x - 1$ will be positive, while $x - 2$ will be negative. For example, if $x = 1.1$ , then $x - 1
   = 0.1$ while $x - 2 = -0.9$ . All together, the fraction $\dfrac{1}{(x - 1)(x - 2)}$ will be negative. But plugging $x = 1$ into the fraction gives $\dfrac{1}{0}$ . Since the result is negative and infinite, it must be $-\infty$ .

You can see numerical evidence for this by plugging (e.g.) $x = 1.001$ into $\dfrac{1}{(x -
   1)(x - 2)}$ .

$$\dfrac{1}{(1.001 - 1)(1.001 - 2)} \approx -1001,$$

a large negative number.

In similar fashion,

$$\lim_{x\to 1^-} \dfrac{1}{(x - 1)(x - 2)} = +\infty,$$

$$\lim_{x\to 2^+} \dfrac{1}{(x - 1)(x - 2)} = +\infty,$$

$$\lim_{x\to 2^-} \dfrac{1}{(x - 1)(x - 2)} = -\infty.$$

Here's the graph:

$$\hbox{\epsfysize=1.75in \epsffile{inflim8.eps}}\quad\halmos$$


Example. The fact that a function is undefined at an isolated value does not imply that it has a vertical asymptote there. For example, $f(x) =
   \dfrac{x^2 - 1}{x - 1}$ is undefined at $x = 1$ . The graph looks like this:

$$\hbox{\epsfysize=1.75in \epsffile{inflim9.eps}}$$

You can see this by noting that, for $x
   \ne 1$ ,

$$\dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1} = x + 1.$$

Thus, the graph is the same as the graph of the line $y = x + 1$ except at $x = 1$ , where there's a hole. In particular, the graph does not have a vertical asymptote at $x = 1$ .


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