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Mathematics LibreTexts

13.6: Limits at Infinity

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    24882
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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.

    Screen Shot 2020-06-25 at 10.15.12 PM.png

    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.

    Screen Shot 2020-06-25 at 10.18.14 PM.png

    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, \(\lim_{x \rightarrow \infty} \frac{sin(x)}{x} = 0\).

    Screen Shot 2020-06-25 at 10.29.23 PM.png

    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.