Limits at Infinity
We recently looked at Limits to Infinity, that is a function $f$ such that $\lim_{x \to a} f(x) = \pm \infty$. We will now look at limits at infinity, that is, as $x \to \pm \infty$, $f(x) \to L$
Definition (Informal); We say that $\lim_{x \to \infty} f(x) = L$ if as $x$ gets larger and larger and tends to $\infty$, then the values $f(x)$ get closer and closer to $L$. Similarly, we say that $\lim_{x \to -\infty} f(x) = L$ if as $x$ gets larger and larger (negatively) and tends to $-\infty$, then the values $f(x)$ get closer and closer to $L$. |
Of course, we will also present the formal definition of a limit at infinity. Once again, this definition is not required for most introductory calculus courses, so feel free to skip over it.
Definition (Formal): Suppose that $f$ is a function defined on an open interval $I = (a, \infty)$. It follows that a limit at positive infinity written $\lim_{x \to \infty} f(x) = L$ means that for every $\epsilon > 0$, there is a $k \in \mathbb{R}$ such that if $x > k$, then $\mid f(x) - L \mid < \epsilon$. Similarly, if $f$ is a function defined on an open interval $I = (-\infty, a)$, it follows that a limit at negative infinity written $\lim_{x \to -\infty} f(x) = L$ means that for every $\epsilon > 0$, there is a $k \in \mathbb{R}$ such that if $x < k$, then $\mid f(x) - L \mid < \epsilon$. |
Essentially our definition says that if we make the distance between $f(x)$ and $L$ arbitrarily small and choose a successively $x > k$ (for limits to positive infinity) or $x < k$ (for limits to negative infinity), then we can find a corresponding get of lines $y = L + \epsilon$ and $y = L - \epsilon$ that we can have both $f(x)$ and $L$ squeeze in between. Clearly, as we choose smaller and smaller $\epsilon$, $\mid f(x) - L \mid$ becomes smaller and smaller, and hence, as $x \to \pm \infty$, $f(x) \to L$.
Example 1
Given the function $f(x) = \frac{1}{x}$, if $x > k$, then $\mid \frac{1}{x} - 0 \mid < 0.5$. Find a sufficient $k$.
We first rewrite the inequality such that $-0.5 < \frac{1}{x} < 0.5$. We note that $L = 0$. If we graph $f(x) = \frac{1}{x}$ as well as the lines $y = 0 + 0.5$ and $y = 0 - 0.5$ together, then:
We note that $f$ intersects the line $y = 0.5$ when $x = 2$. Therefore, for $\epsilon = 0.5$, we can choose any value $k$ such that $k > 2$.