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

(→‎Proof)
(→‎Proof)
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Text explanation:
 
Text explanation:
  
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let <math>z(x) = \frac{f(x)}{g(x)}</math> where <math>f(x)</math> and <math>g(x)</math> are both nonzero function with value <math>0</math> at <math>x = a</math>
+
let <math>z(x) = \frac{f(x)}{g(x)}</math> where <math>f(x)</math> and <math>g(x)</math> are both nonzero functions with value <math>0</math> at <math>x = a</math>
  
 
(for example, <math>g(x) = cos(\frac{\pi}{2} x)</math>, <math>f(x) = 1-x</math>, and <math>a = 1</math>.)
 
(for example, <math>g(x) = cos(\frac{\pi}{2} x)</math>, <math>f(x) = 1-x</math>, and <math>a = 1</math>.)
Line 18: Line 18:
 
Note that the points surrounding z(a) aren't approaching infinity, as a function like <math>f(x) = 1/x-1</math> might at <math>f(a)</math>
 
Note that the points surrounding z(a) aren't approaching infinity, as a function like <math>f(x) = 1/x-1</math> might at <math>f(a)</math>
  
−
The points infinitely close to z(a) will be equal to <math>\lim{b\to \infty} \frac{f(a+b)}{g(a+b)}</math>
+
The points infinitely close to z(a) will be equal to <math>\lim{b\to 0} \frac{f(a+b)}{g(a+b)}</math>
  
−
Noting that <math>\lim{b\to \infty} f(x+b)</math> and <math>\lim{b\to \infty} g(x+b)</math> are equal to <math>f'(x)</math> and <math>g'(x)</math> respectively.  
+
Noting that <math>\lim{b\to 0} f(x+b)</math> and <math>\lim{b\to 0} g(x+b)</math> are equal to <math>f'(x)</math> and <math>g'(x)</math> respectively.  
−
This means that the points approaching <math>\frac{f(x)}{g(x)}</math> at point a where <math>f(a)</math> and <math>g(a)</math> are equal to 0 are equal to $\frac{f'(x)}{g'(x)}
+
This means that the points approaching <math>\frac{f(x)}{g(x)}</math> at point a where <math>f(a)</math> and <math>g(a)</math> are equal to 0 are equal to <math>\frac{f'(x)}{g'(x)}</math>
  
 
==Problems==
 
==Problems==

Revision as of 20:31, 11 March 2022

L'Hopital's Rule is a theorem dealing with limits that is very important to calculus.

Theorem

The theorem states that for real functions $f(x),g(x)$, if $\lim f(x),g(x)\in \{0,\pm \infty\}$ \[\lim\frac{f(x)}{g(x)}=\lim\frac{f'(x)}{g'(x)}\] Note that this implies that \[\lim\frac{f(x)}{g(x)}=\lim\frac{f^{(n)}(x)}{g^{(n)}(x)}=\lim\frac{f^{(-n)}(x)}{g^{(-n)}(x)}\]

Proof

No proof of this theorem is available at this time. You can help AoPSWiki by adding it.

Video by 3Blue1Brown: https://www.youtube.com/watch?v=kfF40MiS7zA

Text explanation:

let $z(x) = \frac{f(x)}{g(x)}$ where $f(x)$ and $g(x)$ are both nonzero functions with value $0$ at $x = a$

(for example, $g(x) = cos(\frac{\pi}{2} x)$, $f(x) = 1-x$, and $a = 1$.)

Note that the points surrounding z(a) aren't approaching infinity, as a function like $f(x) = 1/x-1$ might at $f(a)$

The points infinitely close to z(a) will be equal to $\lim{b\to 0} \frac{f(a+b)}{g(a+b)}$

Noting that $\lim{b\to 0} f(x+b)$ and $\lim{b\to 0} g(x+b)$ are equal to $f'(x)$ and $g'(x)$ respectively. This means that the points approaching $\frac{f(x)}{g(x)}$ at point a where $f(a)$ and $g(a)$ are equal to 0 are equal to $\frac{f'(x)}{g'(x)}$

Problems

Introductory

Intermediate

Olympiad

See Also

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