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

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Note that <math>\lim_{b\to 0} f(a+b)</math> and <math>\lim_{b\to 0} g(a+b)</math> are equal to <math>f'(a)</math> and <math>g'(a)</math>.  
 
Note that <math>\lim_{b\to 0} f(a+b)</math> and <math>\lim_{b\to 0} g(a+b)</math> are equal to <math>f'(a)</math> and <math>g'(a)</math>.  
  
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As a recap, this means that the points approaching <math>\frac{f(a)}{g(a)}</math> where <math>a</math> is a number such that <math>f(a)</math> and <math>g(a)</math> are both equal to 0 are going to approach <math>\frac{f'(x)}{g'(x)}</math>
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As a recap, this means that the points approaching <math>\frac{f(a)}{g(a)}</math>, where <math>a</math> is a number such that <math>f(a)</math> and <math>g(a)</math> are both equal to <math>0</math>, are going to approach <math>\frac{f'(x)}{g'(x)}</math>.
  
 
==Problems==
 
==Problems==

Revision as of 21:23, 13 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, $g(x) = \cos\left(\frac{\pi}{2} x\right)$, $f(x) = 1-x$, 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 $\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}$.

Note that $\lim_{b\to 0} f(a+b)$ and $\lim_{b\to 0} g(a+b)$ are equal to $f'(a)$ and $g'(a)$.

As a recap, this means that the points approaching $\frac{f(a)}{g(a)}$, where $a$ is a number such that $f(a)$ and $g(a)$ are both equal to $0$, are going to approach $\frac{f'(x)}{g'(x)}$.

Problems

Introductory

Intermediate

Olympiad

See Also

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