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# 13.6: Limits at Infinity

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For some functions $$f(x)$$, limits such as $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow -\infty} f(x)$$ make sense. Consider the function graphed below. As x moves to the right (towards positive infinity) the corresponding $$f(x)$$ value approaches 2. We express this in symbols as $$\lim_{x \rightarrow \infty} f(x) = 2$$. Such a limit is called a limit at infinity, which is a bit of a misnomer because x is never “at” infinity, just moving toward it.

The graph squeezes in on the dashed horizontal line $$y = 2$$ as x moves to ∞. This line is called a horizontal asymptote of the function $$f(x)$$. It is not a part of the graph, but it helps us visualize the behavior of $$f(x)$$ as x grows.

Also, in this picture, as x moves to the left (toward negative infinity), the corresponding value $$f(x)$$ approaches $$-1$$. We express this in symbols as $$\lim_{x \rightarrow -\infty} f(x) = -1$$. The horizontal line $$y = -1$$ is a second horizontal asymptote of this function $$f(x)$$.

In general, $$\lim_{x \rightarrow \infty} f(x) = L$$ means that $$f(x)$$ is arbitrarily close to L, provided that x is sufficiently large (i.e., “provided that x is sufficiently close to $$\infty$$”).

In other words, given any $$\epsilon > 0$$, there is a number $$N > 0$$ (possibly quite large) such that $$x > N$$ implies $$|f(x)-L| < \epsilon$$. This is illustrated below.

Analogously, for x approaching $$-\infty$$, we say $$\lim_{x \rightarrow -\infty} f(x) = L$$ means that $$f(x)$$ is arbitrarily close to L, provided x is a sufficiently close to $$-\infty$$. In other words, given any $$\epsilon > 0$$, there is a number $$N < 0$$ such that $$x < N$$ implies $$|f(x)-L| < \epsilon$$. Here is a summary of these ideas.

Definition: Word

1. The statement $$\lim_{x \rightarrow \infty} f(x) = L$$ means that for any real $$\epsilon > 0$$, there is a number $$N > 0$$ for which $$x > N$$ implies $$|f(x)-L| < \epsilon$$.
2. The statement $$\lim_{x \rightarrow -\infty} f(x) = L$$ means that for any real $$\epsilon > 0$$, there is a number $$N < 0$$ for which $$x < N$$ implies $$|f(x)-L| < \epsilon$$.

Example 13.7

Investigate $$\lim_{x \rightarrow \infty} \frac{sin(x)}{x}$$.

For any $$x \in \mathbb{R}$$, we know that $$-1 \le sin(x) \le 1$$. Consequently we would expect $$sin(x)$$ to be very small when x is large, that is, we expect $$\lim_{x \rightarrow \infty} \frac{sin(x)}{x} = 0$$.

Let us use Definition 13.4 to prove this. Given $$\epsilon > 0$$, put $$N = \frac{1}{\epsilon}$$ . If $$x > N$$, then $$x > \frac{1}{\epsilon}$$, so $$\frac{1}{x} < \epsilon$$, and hence $$-\epsilon < \frac{1}{x} sin(x) < \epsilon$$, meaning $$|\frac{sin(x)}{x} < \epsilon$$.

In summary, given $$\epsilon > 0$$, there is an $$N > 0$$ for which $$x > N$$ implies $$|\frac{sin(x)}{x}-0| < \epsilon$$. By Definition 13.4, $$\lim_{x \rightarrow \infty} \frac{sin(x)}{x} = 0$$.

In a similar manner we can prove $$\lim_{x \rightarrow -\infty} \frac{sin(x)}{x} = 0$$. Thus the x-axis $$y = 0$$ is a horizontal asymptote to $$sin(x)$$, as illustrated above.

Of course, not every limit at infinity will exist. Consider $$\lim_{x \rightarrow \infty} x^2$$. As x goes to infinity, the quantity $$x^2$$ approaches infinity too. Common sense says the limit does not exist because $$x^2$$ eventually exceeds any finite number L. But it’s good practice to prove this common-sensical statement.

Suppose for the sake of contradiction that $$\lim_{x \rightarrow -\infty} x^2 = L$$. for some $$L \in \mathbb{R}$$. Let $$\epsilon = 1$$, and apply Definition 13.4 to get a number N for which $$x > N$$ implies $$|x^2-L| < 1$$. The inequality (13.3) yields $$|x^2|-|L| = |x^2|-|-L| \le |x^2+(-L)| = |x^2-L| < 1$$. In other words, $$x^2-|L| < 1$$, or $$x^2 < 1+|L|$$ for all $$x > N$$. But this is false for those x that are bigger than both N and 1 + |L|, a contradiction.

Even though l$$\lim_{x \rightarrow \infty} x^2$$ does not exist, we allow the notation $$\lim_{x \rightarrow \infty} x^2 = \infty$$ to indicate that $$x^2$$ grows without bound as x goes to infinity. In general, $$\lim_{x \rightarrow \infty} x^2 = \infty$$ means that $$f(x)$$ eventually exceeds any number L:

1. $$\lim_{x \rightarrow \infty} x^2 = \infty$$ means that for any real number L, there is a positive N for which $$x > N$$ implies $$f(x) > L$$.
2. $$\lim_{x \rightarrow \infty} x^2 = -\infty$$ means that for any real number L, there is a positive N for which $$x > N$$ implies $$f(x) < L$$.

Limits of the form $$\lim_{x \rightarrow \infty} x^2 = \pm \infty$$ play a small role in the next section.

## Exercise

Use Definition 13.4 to prove the following results. (Where appropriate, you may wish to adapt the corresponding proofs from Section 13.4.)

Exercise $$\PageIndex{1}$$

$$\lim_{x \rightarrow \infty} \frac{1}{x^n} = 0$$ if $$n \in \mathbb{N}$$

Exercise $$\PageIndex{2}$$

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

Exercise $$\PageIndex{3}$$

If $$a \in \mathbb{R}$$,then $$\lim_{x \rightarrow \infty} a = a$$

Exercise $$\PageIndex{4}$$

If $$\lim_{x \rightarrow \infty} f(x)$$ exists, and $$a \in \mathbb{R}$$, then $$\lim_{x \rightarrow \infty} af(x) = a\lim_{x \rightarrow \infty} f(x)$$.

Exercise $$\PageIndex{5}$$

If both $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ exist, then $$\lim_{x \rightarrow \infty} (f(x)+g(x)) = \lim_{x \rightarrow \infty} f(x)+\lim_{x \rightarrow \infty} f(x)$$.

Exercise $$\PageIndex{6}$$

If both $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ exist, then $$\lim_{x \rightarrow \infty} f(x)g(x) = (\lim_{x \rightarrow \infty} f(x) ) \cdot (\lim_{x \rightarrow \infty} f(x))$$.

Exercise $$\PageIndex{7}$$

If both $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ exist, then $$\lim_{x \rightarrow \infty} (f(x)-g(x)) = \lim_{x \rightarrow \infty} f(x)-\lim_{x \rightarrow \infty} f(x)$$.

Exercise $$\PageIndex{8}$$

If both $$\lim_{x \rightarrow \infty} f(x)$$ and $$\lim_{x \rightarrow \infty} g(x)$$ exist, and $$\lim_{x \rightarrow \infty} g(x) \ne 0$$ then $$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow \infty} f(x)}{\lim_{x \rightarrow \infty} f(x)}$$.

Exercise $$\PageIndex{9}$$

If $$\lim_{x \rightarrow \infty} g(x) = L$$ and f is continuous at $$x = L$$, then $$\lim_{x \rightarrow \infty} f(g(x)) = f (\lim_{x \rightarrow \infty} g(x))$$.

Exercise $$\PageIndex{10}$$

Prove that $$\lim_{x \rightarrow \infty} sin(x)$$ does not exist.