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Difference between revisions of "L'Hpital's Rule"
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'''Discovered By'''
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==Discovered By==
  
 
Guillaume de l'Hopital
 
Guillaume de l'Hopital
  
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'''The Rule'''
+
==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.  
  
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'''Examples'''
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==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>
 +
 +
==Extra Help Needed?==
 +
 +
Send a message here: [http://www.artofproblemsolving.com/Forum/ucp.php?i=pm&mode=compose&u=80321

Revision as of 21:54, 15 March 2011

Discovered By

Guillaume de l'Hopital

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

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

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

Extra Help Needed?

Send a message here: [http://www.artofproblemsolving.com/Forum/ucp.php?i=pm&mode=compose&u=80321

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