Let’s now examine the limits at negative infinity of some common functions. We’ll again use epsilon-strips to assist our thinking, and we’ll move faster than we did on the preceding screen since the reasoning is quite similar.
The limit at negative infinity exists and equals L if,
for any value of $\epsilon \gt 0$ that we choose,
there is a value of M such that for all $x \lt M$
the function’s output values lie in the range $L\, – \epsilon \lt f(x) \lt L + \epsilon.$
Note: When looking at functions like $x^n,$ we’re going to restrict ourselves to integer values of n so we don’t have to worry about the nth-root (like $\sqrt{x} = x^{0.5}$) of negative numbers, since we’re now looking at $x \to -\infty.$
What is $\displaystyle{\lim_{x \to -\infty} \sin(x) }?$ How would you explain your conclusion to a friend?
The reasoning and conclusion for what happens as $x \to -\infty$ is the same as that for $x \to \infty$ from the preceding screen:
Conclusion: $\displaystyle{\lim_{x \to -\infty} \sin(x) = \text{ DNE} \, \cmark},$ because there is no value of L for which this works given our chosen $\epsilon = 0.5$ or $=0.1.$
Similarly, $\displaystyle{\lim_{x \to -\infty} \cos(x) = \text{ DNE} \, \cmark}.$
Does $\displaystyle{\lim_{x \to -\infty} e^x }$ exist? If so, what is its value?
As you can see, only the value $L = 0$ satisfies the condition $L – \epsilon \lt f(x) \lt L + \epsilon$ for any $\epsilon \gt 0.$
Conclusion: $\displaystyle{\lim_{x \to -\infty}e^x = 0 \, \cmark}.$
What is $\displaystyle{\lim_{x \to -\infty} x^n }$ for integer values of $n \ge 1$?
You can use the slider beneath the graph to change n. As you try different values, what do you notice about what happens as $x \to -\infty$ for odd versus even values of n?
Currently n = 3: $f(x) = x^3$
Conclusion: $\displaystyle{\lim_{x \to -\infty} x^n } = \text{ DNE} \, \cmark$ for integer $n \ge 1.$ If we choose to convey more information about the way in which the limit does not exist as $x \to -\infty$, we can further specify that
\[ \lim_{x \to -\infty} x^n =
\begin{cases}
\infty, & \text{if $n$ is even} \\[8px]
-\infty, & \text{if $n$ is odd}
\end{cases}
\]
Consider the function $f(x) = C,$ where C is a constant (say, 42).
$\displaystyle{\lim_{x \to -\infty} C = ? }$
Conclusion: Let C be a constant value. Then $\displaystyle{\lim_{x \to -\infty}C = C \, \cmark}$
For $n \ge 1,$ $\displaystyle{\lim_{x \to -\infty}x^{-n} = \lim_{x \to -\infty} \dfrac{1}{x^n} = ? }$
Currently n = 2: $f(x) = x^{-2} = \dfrac{1}{x^2}$
Conclusion: For $n \ge 1,$ $\displaystyle{\lim_{x \to -\infty}x^{-n} = \lim_{x \to -\infty} \dfrac{1}{x^n} = 0 \, \cmark }$
On the next screen we’ll use the conclusions above to lay some groundwork for deciding determining the limit as $x \to \pm\infty$ for many, many more functions.
We’re going to essentially repeat the Upshot from the preceding screen since it still applies:
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