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Calculus Done Right

The mathematics of the continuous, through intuition not memorization.

Infinite Limits

                 

In this quiz, we will explore two types of “infinite limits.” The first type is a bit of a misnomer: in fact, it is an example of a way in which a limit can fail to exist.

Let \(f(x)\) be a function and \(a\) be some real number. We say that \(\lim\limits_{x\to a} f(x) = \infty\) if \(f(x)\) “blows up” as \(x\) gets near \(a\); that is, for all \(x\) sufficiently close to \(a\) (but not equal to \(a\)), \(f(x)\) gets arbitrarily large. When this happens, even though we will write \(\lim\limits_{x\to a} f(x) = \infty,\) we'll still say the limit doesn't exist!

Here’s one way to see what this means: suppose we pick any large number, say \(100000.\) Then there is some interval around \(a\) such that \(f(x) > 100000\) for all the \(x\) in this interval (except possibly for \(x=a\)).

One way to investigate this is by plugging in numbers closer and closer to \(a,\) and seeing that they get larger and larger without bound.

Similarly, \(\lim\limits_{x\to a} f(x) = -\infty\) if \(f(x)\) “blows up” in a downward direction: for all \(x\) sufficiently close to \(a\) (but not equal to \(a\)), \(f(x)\) gets arbitrarily large and negative (i.e. \(-f(x)\) gets arbitrarily large).

Let’s see what \(\lim\limits_{x\to a} f(x) = \infty\) looks like. Which of these is the graph of a function \(f(x)\) such that \(\lim\limits_{x\to 0} f(x) = \infty\)?

When a function is given by a formula instead of a graph, how do we determine if \(\lim\limits_{x\to a} f(x) = \infty\)? We imagine plugging in numbers close to \(a\), and analyze what happens to the expression. We don’t have to actually plug numbers into a calculator; we can just analyze what would happen if \(x\) were close to \(a.\)

For instance, look at \(\dfrac{x+2}{(x-1)^2}\) near \(x=1.\) When \(x\) is close to 1, the denominator is a very small positive number, while the numerator is close to 3. So the whole thing blows up, and so \(\lim\limits_{x\to 1} \dfrac{x+2}{(x-1)^2} = \infty\). Be careful though: If both the numerator and denominator are very small or very large, we have an indeterminate form, and we can’t conclude anything without further work!


Which of these functions satisfies \(\lim\limits_{x\to 1} f(x) = \infty\)?

Remember that writing \(\lim\limits_{x\to a} f(x) = \infty\) is a shorthand for “the limit does not exist because the function blows up at \(a.\)” This shorthand can be useful, but it’s very important not to manipulate it as if \(\infty\) is a number. Consider the following three statements:

  • “\(\infty + \infty = \infty\)”: if \(\lim\limits_{x\to a} f(x) = \infty\) and \(\lim\limits_{x\to a} g(x) = \infty,\) then \(\lim\limits_{x\to a} \left[f(x) + g(x)\right] = \infty.\)
  • “\(\infty - \infty = 0\)”: if \(\lim\limits_{x\to a} f(x) = \infty\) and \(\lim\limits_{x\to a} g(x) = \infty,\) then \(\lim\limits_{x\to a} \left[f(x) - g(x)\right] = 0.\)
  • “\(\infty \cdot \infty = \infty\)”: if \(\lim\limits_{x\to a} f(x) = \infty\) and \(\lim\limits_{x\to a} g(x) = \infty,\) then \(\lim\limits_{x\to a} \left[f(x) \cdot g(x)\right] = \infty.\)

How many of these are true?

The second type of infinite limit is the limit “at infinity”: instead of looking at places where a function \(f(x)\) gets arbitrarily large, we look at what happens to \(f(x)\) as \(x\) gets arbitrarily large. That is, we say that \(\lim\limits_{x\to\infty} f(x) = L\) if \(f(x)\) gets arbitrarily close to \(L\) as \(x\) “approaches \(\infty,\)” i.e. gets arbitrarily large.

So if you pick your favorite small positive number, say \(0.001,\) then \(f(x)\) is within \(0.001\) of \(L\) for all sufficiently large \(x\) (i.e. for all \(x\) in an “interval around \(\infty\)”).

If you picked a large enough \(a,\) you would find that f(x) is within 0.001 of 0 for all \(x>a.\)

If you picked a large enough \(a,\) you would find that f(x) is within 0.001 of 0 for all \(x>a.\)

As with infinite limits, one way to investigate this is by plugging larger and larger values of \(x\) and seeing that \(f(x)\) gets closer and closer to \(L.\)

Similarly we can define \(\lim\limits_{x\to -\infty} f(x)\) by looking at what happens to \(f(x)\) as \(x\) gets arbitrarily large in the negative direction, i.e. farther and farther to the left on the number line.

Now that we know what the limit of \(f(x)\) at infinity means, let’s see what it says about the graph of \(f(x).\)

Which of the following is the graph of a function \(f(x)\) that satisfies \(\lim\limits_{x\to\infty} f(x) = 1\)?

Which of these satisfies \(\lim\limits_{x\to-\infty} f(x) = 0\)?

One family of limits at infinity you will get to know are the reciprocal power functions, like \(\frac{1}{x}\) and \(\frac{1}{x^2}.\) Clearly, \(\lim\limits_{x \to \infty} \frac{1}{x} = 0,\) because as \(x\) gets very large, its reciprocal gets very small. Similarly for \(\frac{1}{x^2}.\)

What about in general? For what values of \(p\) is it the case that

\[\lim_{x \to \infty} \frac{1}{x^p} = 0\ ?\]

You will have to remind yourself what negative and fractional exponents mean!

Consider the following statement:

If \(\lim\limits_{x\to\infty} f(x) = 0,\) then for large enough \(x,\) \(f(x)\) gets closer and closer to \(0\) without ever equaling \(0.\)

Which of the following choices is correct?

A) The statement is true.

B) The statement is false, and \(f(x) = \dfrac{1}{x}\) is a counterexample.

C) The statement is false, and \(f(x) = \dfrac{\ln(x)}{x}\) is a counterexample.

D) The statement is false, and \(f(x) = \dfrac{\sin(x)}{x}\) is a counterexample.

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