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Calculus Done Right

The mathematics of the continuous, through intuition not memorization.

Computing Limits

                     

This quiz will focus on techniques for actually computing limits from algebraic expressions.

To start with, we know \(\lim\limits_{x \to a} f(x)\) for any function that’s continuous at \(a\): it’s just \(f(a).\) How many of the following statements are true?

  • \(\displaystyle \lim_{x \to 1} (x-1)^3 = 0\)

  • \(\displaystyle \lim_{x \to 1} \frac{(x-1)^3}{x} = 0\)

  • \(\displaystyle \lim_{x \to \pi} \cos(x) = -1\)

It’s also true that basic combinations of limits behave as you expect, including sums, products, and quotients. This means, for example:

  • If \(\lim\limits_{x \to a} f(x) = L\), then \(\lim\limits_{x \to a} cf(x) = cL.\)

  • If \(\lim\limits_{x \to a} f(x) = L\) and \(\lim\limits_{x \to a} g(x) = K,\) then \(\lim\limits_{x \to a} \left[f(x) + g(x)\right] = L + K.\)

  • If \(\lim\limits_{x \to a} f(x) = L\) and \(\lim\limits_{x \to a} g(x) = K,\) then \(\lim\limits_{x \to a} f(x)g(x) = LK.\)

Remember the warning from a previous quiz, though: if \(\lim\limits_{x \to a} f(x) = \infty\) and \(\lim\limits_{x \to a} g(x) = \infty\), then while it follows that \(\lim\limits_{x \to a} \left[f(x) + g(x)\right] = \infty,\) you cannot conclude anything about \(\lim\limits_{x \to a} \left[f(x) - g(x)\right].\)

How many of the following statements are true?

  • \(\displaystyle \lim_{x \to 0} \left[\sin(x) + \cos(x)\right] = 1\)

  • \(\displaystyle \lim_{x \to 0}\left[ \left(\frac{x^2 + 1}{x^2} \right) + \left(\frac{1}{x^2}\right)\right] = \infty\)

  • \(\displaystyle \lim_{x \to 0} \left[\left(\frac{x^2 + 1}{x^2} \right) - \left(\frac{1}{x^2}\right)\right] = 0\)

Limits behave with function composition as well, and often the best approach is to think about what’s really happening as the \(x\)’s approach \(a.\) For example, consider \(\lim\limits_{x \to \infty} \cos\left(\frac{1}{x}\right)\). We reason as follows:

As \(x\) gets very large, \(\frac{1}{x}\) approaches 0. What does \(\cos\) do to numbers that are getting close to 0? It sends them to numbers close to 1 (because \(\cos\) is a continuous function and \(\cos(0)=1\)). So the limit is 1.

You should get used to this kind of reasoning about limits. You can memorize properties and manipulate symbols, but this is the way people who deal with limits all the time tend to think about them!

Try this reasoning here. What are

\[\large \lim_{x \to \infty} e^{-\frac{1}{x^2}}\] and \[\large \lim_{x \to 0} e^{-\frac{1}{x^2}}\ ?\]

The following argument is incorrect. At which step is the mistake made?

Claim: \(\lim\limits_{x \rightarrow \infty}\ \left(1+\frac{1}{x}\right)^x = 1\)

Proof:

Step 1: Let \(f(x) = x^x\) and let \(g(x) = 1 + \frac{1}{x}\). We need to compute the limit of the composition, \(\lim\limits_{x \to \infty} f(g(x)),\) and show it equals 1.

Step 2: As \(x\) gets large, \(g(x)\) approaches 1.

Step 3: As \(g(x)\) approaches 1, \(f(g(x))\) also approaches 1, since \(f\) is continuous at 1. QED.

Indeterminate forms have come up before. Let’s look at them carefully now, in the context of computing limits.

When we say “\(\frac{0}{0}\) is an indeterminate form”, we don’t actually mean you divide 0 by 0 (which is undefined). We mean there’s a limit of fractions, and both the numerator and denominator are going to 0. It’s indeterminate because you can’t tell what the limit is from this information--it depends on how fast the top and bottom are getting to 0 relative to each other. It’s like a race.

In contrast--here’s the important point--consider the form “\(\frac{3}{0}\)”. (Again, this doesn’t mean you actually divide by 0. It means there’s a limit of fractions, and the numerator is going to 3, and the denominator is going to 0.) This form is not indeterminate, because it’s perfectly clear what the limit is: As the denominator gets very small and the numerator gets close to 3, the fractions must be getting very large, so the limit here is \(\infty.\)

A form is indeterminate when it doesn’t give you enough information to determine the limit, because the limit depends on how fast quantities are getting to 0 or to \(\infty\) relative to each other.

(One caveat: for the purposes of this kind of analysis, when something goes to 0, we'll assume the values going to 0 are all positive.)

How many of the following forms are indeterminate? This is tricky, and it’s worth thinking about each form carefully. In each case there’s a limit of fractions. In how many of the below cases do you not have enough information to determine the limit?

\[\frac{0}{0} \qquad \frac{10}{0} \qquad \frac{10}{\infty} \qquad \frac{\infty}{\infty} \qquad \frac{0}{\infty} \]

(As a reminder: for the purposes of this kind of analysis, when something goes to 0, we'll assume the values going to 0 are all positive.)

Let’s look in more detail at the \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) indeterminate forms as \(x \to \infty.\) How do we compute these limits? Sometimes we can use algebraic manipulation to convert them to a non-indeterminate form, and calculate the limit.

For example, consider the \(\frac{\infty}{\infty}\) indeterminate form: \[\lim\limits_{x\to\infty} \dfrac{2x^2+x}{x^3-1}.\]

Here is a common trick that works whenever the numerator and denominator are both polynomials: Divide each term on the top and bottom by a specific power of \(x.\) The power of \(x\) to use is the largest one that appears anywhere, in this case \(x^3.\) So:

\[ \begin{align} \lim_{x\to\infty} \frac{2x^2+x}{x^3-1} &= \lim_{x\to\infty} {\large\dfrac{\frac{2x^2}{x^3}+\frac{x}{x^3}}{\frac{x^3}{x^3}-\frac1{x^3}}} \\ \\ &= \lim_{x\to\infty} {\large\dfrac{\frac2{x}+\frac1{x^2}}{1-\frac1{x^3}}}. \end{align} \]

Do you see that this is no longer an indeterminate form? Now, the numerator is a sum of fractions that both approach \(0\) as \(x\) gets large, and the denominator approaches \(1\) as \(x\) gets large, so the fraction approaches \(\frac{0}{1}=0.\) We’ve resolved the ambiguity!

This idea can be used on any rational function, i.e. any function \(f(x)\) of the form \(\dfrac{p(x)}{q(x)}\) where \(p\) and \(q\) are polynomials.

Which of these functions \(f(x)\) does NOT satisfy \(\lim\limits_{x\to\infty} f(x) = 0?\)

Let's summarize the last couple pages into a rule of thumb. Which of the following statements are true? (Use the algebraic trick from the last two pages.)

When evaluating \(\lim\limits_{x \to \infty} \frac{p(x)}{q(x)},\) where \(p\) and \(q\) are polynomials:

I. If \(\text{deg}(p) > \text{deg}(q),\) the limit is always \(\infty\) or \(-\infty.\)

II. If \(\text{deg}(p) < \text{deg}(q),\) the limit is always 0.

III. If \(\text{deg}(p) = \text{deg}(q),\) the limit is always the ratio of the leading coefficients.

There’s one more rule of thumb to consider: exponentials always dominate polynomials as \(x \to \infty.\) (We’ll see why this is true in Chapter 4.) For example, consider

\[\lim_{x \to \infty} \frac{100000x^{100000}}{2^x}\]

This is another \(\frac{\infty}{\infty}\) indefinite form. We can’t transform it by algebraic manipulation, but we can use the rule of thumb to say that even though the numerator involves large powers of x, the denominator is an exponential function, so it will approach \(\infty\) faster. This limit is 0.

Which of these functions does not have a limit of 0 as \(x \to \infty\)?

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