Day 9 in Mat229

Remember that the limit definition of the derivative is itself indeterminate: \[ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \]

Suppose $h(x)=\frac{f(x)}{g(x)}$, where $f$ and $g$ are differentiable and $g'(x) \ne 0$ near $x=a$ (except possibly at $a).$ Suppose that \[ \lim_{x \to a}f(x)=0 \;\;{\text{and}}\;\; \lim_{x \to a}g(x)=0 \] Then \[ \lim_{x \to a}h(x)= \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \] This holds assuming the limit on the right exists or is \(\infty\) or \(-\infty\). This result also holds if we are considering one-sided limits, or if \(a\) is replaced by \(\infty\) or \(-\infty\).

\[ \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} = \frac{f'(a)}{g'(a)} \] 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" \[ \begin{align*} \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)} = \lim_{x \to a} \frac{f'(a)}{g'(a)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \end{align*} \]

• We saw an example of this type yesterday, in Example 4.21:
  

  A function that appears well-behaved around \(x=0\), even though it's not defined there. So while it appears that \[ \lim_{x \to 0} \frac{\sin(x)}{x} \]

  exists, and is in fact 1, how do we know that? L'Hôpital's rule gives us the tools to figure it out.

  This function is so important that we have defined a piecewise function to heal its hole: \[ \rm{sinc}(x) \equiv \left\{\begin{array}{cc} {\frac{\sin(x)}{x}}&{x \ne 0}\cr {1}&{x=0} \end{array}\right. \]

  It's used a lot in image processing and interpolation.

• As noted in our text the rule also works for the case where \(f\) and \(g\) to to \(\infty\) or \(-\infty\) as well (Theorem 4.13).

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

  5. $\displaystyle \lim_{x \to \infty} \frac{\ln(x)}{5x}$

• If L'Hôpital's rule is not appropriate for a limit above, does L'Hôpital's rule give the right answer anyway? Try it on any that fail to be appropriate for L'Hôpital's rule.

\(1^\infty\), \(0 * \infty\), \(\infty^0\), \(\infty - \infty\), etc.

They may require other tricks (sometimes just algebra!) to figure out. For examples,

• let's consider a case where a log comes in handy, Exercise 394 (this case is treated in general in Checkpoint 4.41: \[ \lim_{x \to 0^-}\left(1-\frac{1}{x}\right)^x \]

• A really important case where we can use this trick is for compound interest. Let's investigate this quantity (note that L'Hôpital's rule works for infinite limits, too): \[ \lim_{n \to \infty}\left(1+\frac{r}{n}\right)^{nt} \]

The first part of the example also illustrates a standard trick when using L'Hôpital's rule: sometimes you have to use it more than once. The first application may leave you with another indeterminate limit -- so do it again! f

The upshot of this example is that "\(e^x\) grows more rapidly than \(x^p\) for any \(p>0\)", and that "\(x^p\) grows more rapidly than \(\ln(x)\) for any \(p>0\)".