Day 9 in Mat229

Remember that the limit definition of the derivative is itself indeterminate: $$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$$

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

$$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}=\frac{f^{\prime }(a)}{g^{\prime }(a)}$$ provided the derivatives on the right side exists, and $g^{\prime }(a)\ne 0$.

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 $$\underset{x\to a}{lim}h(x)=\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f(x)-f(a)}{g(x)-g(a)}=\underset{x\to a}{lim}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}}=\frac{\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}}{\underset{x\to a}{lim}\frac{g(x)-g(a)}{x-a}}=\frac{f^{\prime }(a)}{g^{\prime }(a)}$$

So $$\begin{gathered} \underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{L_f(x)}{L_g(x)} \\ =\underset{x\to a}{lim}\frac{f^{\prime }(a)(x-a)+f(a)}{g^{\prime }(a)(x-a)+g(a)}\text{.} \end{gathered}$$ "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" $$\begin{array}{r}\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(a)(x-a)}{g^{\prime }(a)(x-a)}=\underset{x\to a}{lim}\frac{f^{\prime }(a)}{g^{\prime }(a)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)}\end{array}$$

• 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 $$\underset{x\to 0}{lim}\frac{\sin (x)}{x}$$

  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: $$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(\mathrm{x})\equiv \{\begin{array}{cc}\frac{\sin (x)}{x}& x\ne 0\\ 1& x=0\end{array}$$

  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 $\mathrm{\infty }$ or $-\mathrm{\infty }$ 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. ${\displaystyle \underset{x\to 0}{lim}\frac{\ln (1+x)}{x}}$

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

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

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

  5. ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{\ln (x)}{5x}}$

• 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.

$1^{\mathrm{\infty }}$, $0\ast \mathrm{\infty }$, ${\mathrm{\infty }}^0$, $\mathrm{\infty }-\mathrm{\infty }$, 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: $$\underset{x\to 0^{-}}{lim}{(1-\frac{1}{x})}^x$$

• 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): $$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{r}{n})}^{nt}$$

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 $p>0$".