Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "L'Hpital's Rule"

(l'Hopital's Rule article)
 
(→‎Discovered By)
 
(6 intermediate revisions by one other user not shown)
Line 1: Line 1:
−
[b][size=150]Discovered by[/size][/b]
+
==Discovered By==
  
−
Guillaume de l'Hopital
+
Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli.
  
−
[b][size=150]The Rule[/size][/b]
+
==The Rule==
  
 
For <math> \frac {0}{0} </math> or <math> \frac {\infty}{\infty} </math> case, the limit <cmath> \lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)} </cmath> where <math> f'(x) </math> and <math> g'(x) </math> are the first derivatives of <math> f(x) </math> and <math> g(x) </math>, respectively.  
 
For <math> \frac {0}{0} </math> or <math> \frac {\infty}{\infty} </math> case, the limit <cmath> \lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)} </cmath> where <math> f'(x) </math> and <math> g'(x) </math> are the first derivatives of <math> f(x) </math> and <math> g(x) </math>, respectively.  
  
−
[b][size=150]Examples[/size][/b]
+
==Examples==
  
 
<math> \lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48} </math>
 
<math> \lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48} </math>
  
 
<math> \lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1} </math>
 
<math> \lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1} </math>
 +
 +
==Proof of l'Hôpital's rule==
 +
A standard proof of l'Hôpital's rule uses [[Cauchy's mean value theorem]]. l'Hôpital's rule has many variations depending on whether ''c'' and ''L'' are finite or infinite, whether ''f'' and ''g'' converge to zero or infinity, and whether the limits are one-sided or two-sided.
 +
===Zero over zero===
 +
Suppose that ''c'' and ''L'' are finite and ''f'' and ''g'' converge to zero.
 +
 +
First, define (or redefine) ''f''(''c'')&nbsp;=&nbsp;0 and ''g''(''c'')&nbsp;=&nbsp;0. This makes ''f'' and ''g'' continuous at ''c'', but does not change the limit (since, by definition, the limit does not depend on the value at the point ''c'').
 +
 +
Since <math>\lim_{x\to c}\frac {f'(x)}{g'(x)}</math> exists, there is an interval (''c''&nbsp;&minus;&nbsp;''δ'',&nbsp;''c''&nbsp;+&nbsp;''δ'') such that for all ''x'' in the interval, with the possible exception of ''x''&nbsp;=&nbsp;''c'', both <math> f'(x)</math> and <math>g'(x)</math> exist and <math>g'(x)</math> is not zero.
 +
 +
If ''x'' is in the interval (''c'',&nbsp;''c''&nbsp;+&nbsp;''δ''), then the [[mean value theorem]] and Cauchy's mean value theorem both apply to the interval [''c'',&nbsp;''x''] (and a similar statement holds for ''x'' in the interval (''c''&nbsp;&minus;&nbsp;''δ'',&nbsp;''c'')). The mean value theorem implies that ''g''(''x'') is not zero (since otherwise there would be a ''y'' in the interval (''c'',&nbsp;''x'') with <math>g'(y)=0</math>). Cauchy's mean value theorem now implies that there is a point ''ξ''<sub>''x''</sub> in (''c'',&nbsp;''x'') such that
 +
 +
:<math>\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.</math>
 +
 +
If ''x'' approaches ''c'', then ''ξ''<sub>''x''</sub> approaches ''c'' (by the [[squeeze theorem]] ).  Since <math>\lim_{x\to c}f'(x)/g'(x)</math> exists, it follows that
 +
 +
:<math>
 +
\lim_{x\to c}\frac{f(x)}{g(x)}
 +
= \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)}
 +
= \lim_{x\to c}\frac{f'(x)}{g'(x)}.
 +
</math>
 +
 +
===Infinity over infinity===
 +
Suppose that ''L'' is finite, ''c'' is positive infinity, and ''f'' and ''g'' converge to positive infinity.
 +
 +
For every ''ε''&nbsp;>&nbsp;0, there is an ''m'' such that
 +
 +
:<math>\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.</math>
 +
 +
The mean value theorem implies that if ''x''&nbsp;>&nbsp;''m'', then ''g''(''x'')&nbsp;≠&nbsp;''g''(''m'') (since otherwise there would be a ''y'' in the interval (''m'',&nbsp;''x'') with <math>g'(y)=0</math>). [[Cauchy's mean value theorem]] applied to the interval [''m'',&nbsp;''x''] now implies that
 +
 +
:<math>\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.</math>
 +
 +
Since ''f'' converges to positive infinity, if ''x'' is large enough, then ''f''(''x'')&nbsp;≠&nbsp;''f''(''m''). Write
 +
 +
:<math>\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.</math>
 +
 +
Now,
 +
 +
:<math>
 +
\begin{align}
 +
& \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\
 +
& \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\
 +
& \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|.
 +
\end{align}
 +
</math>
 +
 +
For ''x'' sufficiently large, this is less than ''ε'' and therefore
 +
 +
:<math>\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.</math>*
 +
* Note: Steps are missing.

Latest revision as of 22:10, 2 January 2012

Discovered By

Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli.

The Rule

For $\frac {0}{0}$ or $\frac {\infty}{\infty}$ case, the limit \[\lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)}\] where $f'(x)$ and $g'(x)$ are the first derivatives of $f(x)$ and $g(x)$, respectively.

Examples

$\lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48}$

$\lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1}$

Proof of l'Hôpital's rule

A standard proof of l'Hôpital's rule uses Cauchy's mean value theorem. l'Hôpital's rule has many variations depending on whether c and L are finite or infinite, whether f and g converge to zero or infinity, and whether the limits are one-sided or two-sided.

Zero over zero

Suppose that c and L are finite and f and g converge to zero.

First, define (or redefine) f(c) = 0 and g(c) = 0. This makes f and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c).

Since $\lim_{x\to c}\frac {f'(x)}{g'(x)}$ exists, there is an interval (c − δ, c + δ) such that for all x in the interval, with the possible exception of x = c, both $f'(x)$ and $g'(x)$ exist and $g'(x)$ is not zero.

If x is in the interval (c, c + δ), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [c, x] (and a similar statement holds for x in the interval (c − δ, c)). The mean value theorem implies that g(x) is not zero (since otherwise there would be a y in the interval (c, x) with $g'(y)=0$). Cauchy's mean value theorem now implies that there is a point ξx in (c, x) such that

$\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.$

If x approaches c, then ξx approaches c (by the squeeze theorem ). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, it follows that

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

Infinity over infinity

Suppose that L is finite, c is positive infinity, and f and g converge to positive infinity.

For every ε > 0, there is an m such that

$\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.$

The mean value theorem implies that if x > m, then g(x) ≠ g(m) (since otherwise there would be a y in the interval (m, x) with $g'(y)=0$). Cauchy's mean value theorem applied to the interval [m, x] now implies that

$\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.$

Since f converges to positive infinity, if x is large enough, then f(x) ≠ f(m). Write

$\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.$

Now,

$\begin{align}

& \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align}$ (Error compiling LaTeX. Unknown error_msg)

For x sufficiently large, this is less than ε and therefore

$\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.$*
  • Note: Steps are missing.
Retrieved from ""