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

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Note that this implies that <cmath>\lim\frac{f(x)}{g(x)}=\lim\frac{f^{(n)}(x)}{g^{(n)}(x)}=\lim\frac{f^{(-n)}(x)}{g^{(-n)}(x)}</cmath>
 
Note that this implies that <cmath>\lim\frac{f(x)}{g(x)}=\lim\frac{f^{(n)}(x)}{g^{(n)}(x)}=\lim\frac{f^{(-n)}(x)}{g^{(-n)}(x)}</cmath>
 
==Proof==
 
==Proof==
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:''No proof of this theorem is available at this time. You can help AoPSWiki by [http://www.artofproblemsolving.com/Wiki/index.php?title=L%27Hopital%27s_Rule&action=edit adding it].''
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One can prove using linear approximation:
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The definition of a derivative is <math>f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}</math> which can be rewritten as <math>f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)</math>.
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Just so all of us know <math>\eta(h)</math> is a function that is both continuous and has a limit of <math>0</math> as the <math>h</math> in the derivative function approaches <math>0</math>.
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After multiplying the equation above by <math>h</math>, we get <math>f(x+h) = f(x) +f'(x)h+h\cdot \eta(h)</math>.
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We have already assumed by the hypothesis that the derivative equals zero. Hence, we can rewrite the function as <math>\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}</math>, which would hence prove our lemma for L'Hospital's rule.
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 +
 
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Video by 3Blue1Brown: https://www.youtube.com/watch?v=kfF40MiS7zA
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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 functions with value <math>0</math> at <math>x = a</math>.
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(For example, <math>g(x) = \cos\left(\frac{\pi}{2} x\right)</math>, <math>f(x) = 1-x</math>, and <math>a = 1</math>.)
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Note that the points surrounding <math>z(a)</math> aren't approaching infinity, as a function like <math>f(x) = 1/x-1</math> might at <math>f(a)</math>.
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The points infinitely close to <math>z(a)</math> will be equal to <math>\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}</math>.
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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>.
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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==
 
===Introductory===
 
===Introductory===
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*Evaluate the limit <math>\lim_{x\rightarrow3}\frac{x^{2}-4x+3}{x^{2}-9}</math> (<url>weblog_entry.php?t=168186 Source</url>)
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*Evaluate the limit <math>\lim_{x\rightarrow3}\frac{x^{2}-4x+3}{x^{2}-9}</math> ([[weblog_entry.php?t=168186 Source]])
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===Intermediate===
 
===Intermediate===
 
===Olympiad===
 
===Olympiad===

Latest revision as of 14:04, 24 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

One can prove using linear approximation: The definition of a derivative is $f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$ which can be rewritten as $f'(x) = \frac{f(x+h)-f(x)}{h} + \eta(h)$. Just so all of us know $\eta(h)$ is a function that is both continuous and has a limit of $0$ as the $h$ in the derivative function approaches $0$. After multiplying the equation above by $h$, we get $f(x+h) = f(x) +f'(x)h+h\cdot \eta(h)$.

We have already assumed by the hypothesis that the derivative equals zero. Hence, we can rewrite the function as $\frac{f(x_0+h)}{g(x_0+h)}=\frac{f'(x_0)h + h\cdot \eta(h)}{g'(x_0)h+h\cdot \epsilon(h)}$, which would hence prove our lemma for L'Hospital's rule.


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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