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

Revision as of 20:18, 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, $g(x) = \cos(\frac{\pi}{2} x)$ , $f(x) = 1-x$ , 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 $\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}$ .

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

Evaluate the limit $\lim_{x\rightarrow3}\frac{x^{2}-4x+3}{x^{2}-9}$ (weblog_entry.php?t=168186 Source)

@@ Line 12: / Line 12: @@
 Text explanation:
-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>
+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>.)
-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 <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>.
-The points infinitely close to z(a) will be equal to <math>\lim{b\to 0} \frac{f(a+b)}{g(a+b)}</math>
+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>.
-Noting 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>.
 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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Difference between revisions of "L'Hôpital's Rule"

Revision as of 20:18, 13 March 2022

Contents

Theorem

Proof

Problems

Introductory

Intermediate

Olympiad

See Also