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Suppose that f(a)=g(a)=0 and that f and g are differentiable on an open interval <i>I</i> containing a.  Suppose also that g'(x)/=0 on <i>I</i> if x/=a.  Then
 
Suppose that f(a)=g(a)=0 and that f and g are differentiable on an open interval <i>I</i> containing a.  Suppose also that g'(x)/=0 on <i>I</i> if x/=a.  Then
 
<math>
 
<math>
\lim_{x -> a}\frac{f(x)}{g(x)}= \lim_{x-> a}\frac{f'(x)}{g'(x)}
+
\lim_{x \to\ a}\frac{f(x)}{g(x)}= \lim_{x \to\ a}\frac{f'(x)}{g'(x)}
 
</math>,
 
</math>,
 
if the limis on the right exists (or is positive or negative infinity).
 
if the limis on the right exists (or is positive or negative infinity).
  
 
This is Elizabeth's favorite theorem.
 
This is Elizabeth's favorite theorem.
 
 
<math>
 
\displaystyle\lim_{x\to\a}\frac{f(x)}{g(x)}=\displaystyle\lim_{x\to\a}\frac{f'(x)}{g'(x)}
 
</math>,
 

Revision as of 11:41, 4 September 2008

Suppose that f(a)=g(a)=0 and that f and g are differentiable on an open interval I containing a. Suppose also that g'(x)/=0 on I if x/=a. Then $ \lim_{x \to\ a}\frac{f(x)}{g(x)}= \lim_{x \to\ a}\frac{f'(x)}{g'(x)} $, if the limis on the right exists (or is positive or negative infinity).

This is Elizabeth's favorite theorem.

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010