Day 20, Mat128: Calculus A

• My office hour Zoom for this afternoon will move to 5:00 because of a meeting.

• If you need to Zoom to class, here's the link.

• We're now moving away from how to COMPUTE derivatives to how to USE derivatives.

  They originated as a way for us to measure rates of change. We will be using that to identify asymptotes, discover behavior of functions near problem points (indeterminate limits), and find the locations of maxima and minima in particular (as part of our optimization units).

  In addition we know that second derivatives tell us about concavity -- so we're close to being able to tell a lot about "the shape of a curve" by using this tool.

I've posted a couple of keys to worksheets (not all have been graded yet -- I'm getting caught up slowly, however!):

1. Section 2.7: Chain Rule/Implicit Differentiation Worksheet

   This one was a little tricky, and you'll want to make sure that you can get the answers. If not, pop in at an office hour for some help.

2. "Chapter 2" worksheet (Power, Quotient, Exponential, and Sine/Cosine Worksheet)
3. Sections 2.1 and 2.3: Power and Linear Rules Worksheet

Notice how I write "Hôpital": it was evidently originally spelled like a large dispensory of societal healthcare, but I don't like using that form because it may make mathematicians pronounce it incorrectly.... So "L'Hôpital" it is!

• Let's begin with the preview example, because it shows off the linearization quite nicely. The functions are \[ h(x)=\frac{f(x)}{g(x)} \] where $f(x)=x^5+x-2$, and $g(x)=x^2-1$. We can see what the problem is: both $f$ and $g$ have roots at $x=1$, resulting in a hole at $x=1$ for $h$ (because we can't compute the quotient there):

  But we can see what's happening around $x=1$, by the linearizations of $f$ and $g$ in the vicinity of $x=1$:

• Compute the linearizations:
  1. $lf(x)=$
  2. $lg(x)=$

• Now, as we get arbitrarily close to $x=1$, we see that we can replace the functions $f$ and $g$ with these linearizations. Hence, in the limit \[ \lim_{x \rightarrow 1}h(x)=\lim_{x \rightarrow 1}\frac{f(x)}{g(x)} \] we replace $f$ and $g$ by their linearizations, and we get some nice cancellation: \[ \lim_{x \rightarrow 1}h(x)=\lim_{x \rightarrow 1}\frac{lf(x)}{lg(x)} =\lim_{x \rightarrow 1}\frac{6(x-1)}{2(x-1)} =\lim_{x \rightarrow 1}\frac{6}{2} =\frac{f'(1)}{g'(1)} =3 = \lim_{x \to 1}\frac{f'(x)}{g'(x)} \] Just as we suspected, from the graph above!

• And this result can be generalized, into "L'Hôpital's rule":

  Suppose $h(x)=\frac{f(x)}{g(x)}$, where $f$ and $g$ are differentiable and $g'(x) \ne 0$ near $a$ (except possibly at $a$). Suppose that \[ \lim_{x \to a}f(x)=0 \;\;{\text{and}}\;\; \lim_{x \to a}g(x)=0 \] Then \[ \lim_{x \to a}h(x)= \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \] if the limit on the right side exists.

  Now let's see how to derive this (or to make sense of it, at any rate): since $f(a)=g(a)=0$, we can write \[ \lim_{x \to a}h(x) = \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f(x)-f(a)}{g(x)-g(a)} = \lim_{x \to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{\lim_{x \to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x \to a}\frac{g(x)-g(a)}{x-a}} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \]

• Examples:
  • Activity 2.8.2. Evaluate each of the following limits. If you use L'Hôpital's rule, indicate where it was used, and be certain its hypotheses are met before you apply it:
    1. $\displaystyle \lim_{x \to 0} \frac{\ln(1 + x)}{x}$

    2. $\displaystyle \lim_{x \to \pi} \frac{\cos(x)}{x}$

    3. $\displaystyle \lim_{x \to 1} \frac{2 \ln(x)}{1-e^{x-1}}$

    4. $\displaystyle \lim_{x \to 0} \frac{\sin(x) - x}{\cos(2x)-1}$

  • Activity 2.8.3: In this activity, we reason graphically from the following figure to evaluate limits of ratios of functions about which some information is known.
    

• The upshot of part c. above is that we may have to use L'Hôpital's rule more than once, if the ratio of the derivatives is also indeterminate. We'll see that in our final example below.

• Finally, we observe that L'Hôpital's rule can even be used in indeterminate cases involving $x \to \infty$, and/or limits like $\frac{\infty}{\infty}$. So, for example, we can find \[ \lim_{x \to \infty} \frac{x^2}{e^x} \] \[ \lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0 \] This function has a horizontal asymptote of $y=0$ as $x \to \infty$:

• Here's a Section 2.8: using derivatives to Evaluate limits worksheet for us to work on (a bit of "review" of the inverse trig functions, too -- let's all take a look at those together, to start).

• A Chain Rule tutorial
• Our free, on-line textbook, called Active Calculus.