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== | ||
− | :'' | + | One can prove using linear approximation: |
+ | 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>. | ||
+ | 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>. | ||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | |||
Video by 3Blue1Brown: https://www.youtube.com/watch?v=kfF40MiS7zA | Video by 3Blue1Brown: https://www.youtube.com/watch?v=kfF40MiS7zA | ||
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Text explanation: | 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>. | |
+ | |||
+ | (For example, <math>g(x) = \cos\left(\frac{\pi}{2} x\right)</math>, <math>f(x) = 1-x</math>, and <math>a = 1</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 <math>z(a)</math> will be equal to <math>\lim_{b\to 0} \frac{f(a+b)}{g(a+b)}</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 <math>0</math>, are going to approach <math>\frac{f'(x)}{g'(x)}</math>. | |
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==Problems== | ==Problems== |
Latest revision as of 15:04, 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 , if
Note that this implies that
Proof
One can prove using linear approximation:
The definition of a derivative is which can be rewritten as
.
Just so all of us know
is a function that is both continuous and has a limit of
as the
in the derivative function approaches
.
After multiplying the equation above by
, we get
.
We have already assumed by the hypothesis that the derivative equals zero. Hence, we can rewrite the function as , which would hence prove our lemma for L'Hospital's rule.
Video by 3Blue1Brown: https://www.youtube.com/watch?v=kfF40MiS7zA
Text explanation:
Let , where
and
are both nonzero functions with value
at
.
(For example, ,
, and
.)
Note that the points surrounding aren't approaching infinity, as a function like
might at
.
The points infinitely close to will be equal to
.
Note that and
are equal to
and
.
As a recap, this means that the points approaching , where
is a number such that
and
are both equal to
, are going to approach
.
Problems
Introductory
- Evaluate the limit
(weblog_entry.php?t=168186 Source)