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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 <math>f(a)=g(a)=0</math> and that f and g are differentiable on an open interval <i>I</i> containing a. <br>
 +
Suppose also that <math>g'(x)\neq0</math> on <i>I</i> if <math>x\neq a</math>. <br>
 +
Then <br>
 
<math>
 
<math>
\displaystyle\lim_{x\to\a}\frac{f(x)}{g(x)}=\displaystyle\lim_{x\to\a}\frac{f'(x)}{g'(x)}
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\lim_{x \to\ a}\frac{f(x)}{g(x)}= \lim_{x \to\ a}\frac{f'(x)}{g'(x)}
</math>,
+
</math>, <br>
if the limis on the right exists (or is positive or negative infinity).
+
if the limit on the right exists (or is <math>\infty</math> or -<math>\infty</math>
 +
).
  
 
This is Elizabeth's favorite theorem.
 
This is Elizabeth's favorite theorem.

Latest revision as of 12:50, 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)\neq0 $ on I if $ x\neq a $.
Then
$ \lim_{x \to\ a}\frac{f(x)}{g(x)}= \lim_{x \to\ a}\frac{f'(x)}{g'(x)} $,
if the limit on the right exists (or is $ \infty $ or -$ \infty $ ).

This is Elizabeth's favorite theorem.

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