Day 33, Mat128: Calculus A

• It will cover from 2.3 through 2.7. I include 2.3 to emphasize the importance of the derivative function.

• You may expect to see another limit definition problem on the exam. Do not forget the limit definition! It's the key to everything else.

• You may also expect to see a "proof/demonstration" of something.

• You should also expect to see some problems that look just like your homework.

Here's a key.

So once we have \(y'\) in terms of \(x\) and \(y\), we can sub in and compute the slope of the tangent at that point.

Knowing the point and the slope, we can write the equation of the tangent line.

• 2a
• 4b

And here is a key for that worksheet.

This is a technique for evaluating indeterminate limits. Remember that the limit definition of the derivative is indeterminate.

Note: We cannot, however, use this technique to resolve that particular case, because what we get is the rather useless "\(f'(x)=f'(x)\)" :)

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 \[ \lim_{x \to a}h(x)= \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)} \] provided the derivatives on the right side exists, and $g'(a) \ne 0$.

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 \[ \lim_{x \to a}h(x) = \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f(x)-f(a)}{g(x)-g(a)} = \lim_{x \to a}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{\lim_{x \to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x \to a}\frac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)} \]

So \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{L_f(x)}{L_g(x)}\\ = \lim_{x \to a} \frac{f'(a)(x-a) + f(a)}{g'(a)(x-a) + g(a)}\text{.} \] "Next, we remember that both \(f(a)=0\) and \(g(a)=0\), which is precisely what makes the original limit indeterminate. Substituting these values for \(f(a)\) and \(g(a)\) in the limit above, we now have" \[ \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)}= \lim_{x \to a} \frac{f'(a)}{g'(a)}\text{,} \end{align*} \]

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