l'H\^opital's rule LHpitalsRule 2002-02-25 00:40:23 2004-04-07 17:37:49 Theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here L'H\^opital's rule states that given an unresolvable limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, the ratio of functions $\frac{f(x)}{g(x)}$ will have the same limit at $c$ as the ratio $\frac{f'(x)}{g'(x)}$. In short, if the limit of a ratio of functions approaches an indeterminate form, then $$\lim_{x\rightarrow c}\frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$$ provided this last limit exists. L'H\^opital's rule may be applied indefinitely as long as the conditions are satisfied. However it is important to note, that the nonexistance of $\lim\frac{f'(x)}{g'(x)}$ does not prove the nonexistance of $\lim\frac{f(x)}{g(x)}$. \textbf{Example:} We try to determine the value of $$\lim_{x\to \infty}\frac{x^2}{e^x}.$$ As $x$ approaches $\infty$ the expression becomes an indeterminate form $\frac{\infty}{\infty}$. By applying L'H\^opital's rule twice we get $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to \infty}\frac{2x}{e^x}=\lim_{x\to \infty}\frac{2}{e^x}=0.$$ Another example of the usage of L'H\^opital's rule can be found \PMlinkid{here}{5741}. %\textbf{Proof of L'H\^opital's Rule for} $\frac{0}{0}$ : %Let there be two differentiable functions $f(x)$ and $g(x)$, such that trying %to evaluate $\lim_{x\rightarrow c}\frac{f(x)}{g(x)}$ yields an indeterminate %form $\frac{0}{0}$. The linearization of $f$ at $c$ is $f(x)\approx %f(c)+f'(c)(x-c)$, and is similar for $g(x)$. Since $x$ is approaching $c$, we %can use the linearization at $c$ to replace $f(x)$ and $g(x)$. Thus %$$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow %c}\frac{f(c)+f'(c)(x-c)}{g(c)+g'(c)(x-c)}$$ %We know that $f(c)$ and $g(c)$ approach $0$, so we can ignore these terms. The %$x-c$ term cancels out and we're left with %$\frac{f'(c)}{g'(c)}$. %$\displaystyle \lim_{x\rightarrow c}\frac{f(x)}{g(x)} = %\lim_{x\rightarrowc}\frac{f'(x)}{g'(x)}$ %Another way to prove this statement is by using Taylor's theorem, instead of %linearization. Thus %$$f(x) = f'(c)(x-c) + R(x)(x-c)^2$$ %where $R(x)$ is a continuous function. When we use this substitution in the %limit, we see %$$\lim_{x\rightarrow c} \frac{f'(c)(x-c) + R(x)(x-c)^2}{g'(c)(x-c) + %S(x)(x-c)^2}$$ %We can factor out an $x-c$ term leaving %$$\lim_{x\rightarrow c} \frac{f'(c) + R(x)(x-c)}{g'(c) + S(x)(x-c)}$$ %which approaches $\frac{f'(c)}{g'(c)}$. %[examples and proof at infinity coming]