Difference between revisions of "L'Hpital's Rule"
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− | + | ==Discovered By== | |
− | Guillaume de l'Hopital | + | Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli. |
− | + | ==The Rule== | |
For <math> \frac {0}{0} </math> or <math> \frac {\infty}{\infty} </math> case, the limit <cmath> \lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)} </cmath> where <math> f'(x) </math> and <math> g'(x) </math> are the first derivatives of <math> f(x) </math> and <math> g(x) </math>, respectively. | For <math> \frac {0}{0} </math> or <math> \frac {\infty}{\infty} </math> case, the limit <cmath> \lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)} </cmath> where <math> f'(x) </math> and <math> g'(x) </math> are the first derivatives of <math> f(x) </math> and <math> g(x) </math>, respectively. | ||
− | + | ==Examples== | |
<math> \lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48} </math> | <math> \lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48} </math> | ||
<math> \lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1} </math> | <math> \lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1} </math> | ||
+ | |||
+ | ==Proof of l'Hôpital's rule== | ||
+ | A standard proof of l'Hôpital's rule uses [[Cauchy's mean value theorem]]. l'Hôpital's rule has many variations depending on whether ''c'' and ''L'' are finite or infinite, whether ''f'' and ''g'' converge to zero or infinity, and whether the limits are one-sided or two-sided. | ||
+ | ===Zero over zero=== | ||
+ | Suppose that ''c'' and ''L'' are finite and ''f'' and ''g'' converge to zero. | ||
+ | |||
+ | First, define (or redefine) ''f''(''c'') = 0 and ''g''(''c'') = 0. This makes ''f'' and ''g'' continuous at ''c'', but does not change the limit (since, by definition, the limit does not depend on the value at the point ''c''). | ||
+ | |||
+ | Since <math>\lim_{x\to c}\frac {f'(x)}{g'(x)}</math> exists, there is an interval (''c'' − ''δ'', ''c'' + ''δ'') such that for all ''x'' in the interval, with the possible exception of ''x'' = ''c'', both <math> f'(x)</math> and <math>g'(x)</math> exist and <math>g'(x)</math> is not zero. | ||
+ | |||
+ | If ''x'' is in the interval (''c'', ''c'' + ''δ''), then the [[mean value theorem]] and Cauchy's mean value theorem both apply to the interval [''c'', ''x''] (and a similar statement holds for ''x'' in the interval (''c'' − ''δ'', ''c'')). The mean value theorem implies that ''g''(''x'') is not zero (since otherwise there would be a ''y'' in the interval (''c'', ''x'') with <math>g'(y)=0</math>). Cauchy's mean value theorem now implies that there is a point ''ξ''<sub>''x''</sub> in (''c'', ''x'') such that | ||
+ | |||
+ | :<math>\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.</math> | ||
+ | |||
+ | If ''x'' approaches ''c'', then ''ξ''<sub>''x''</sub> approaches ''c'' (by the [[squeeze theorem]] ). Since <math>\lim_{x\to c}f'(x)/g'(x)</math> exists, it follows that | ||
+ | |||
+ | :<math> | ||
+ | \lim_{x\to c}\frac{f(x)}{g(x)} | ||
+ | = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} | ||
+ | = \lim_{x\to c}\frac{f'(x)}{g'(x)}. | ||
+ | </math> | ||
+ | |||
+ | ===Infinity over infinity=== | ||
+ | Suppose that ''L'' is finite, ''c'' is positive infinity, and ''f'' and ''g'' converge to positive infinity. | ||
+ | |||
+ | For every ''ε'' > 0, there is an ''m'' such that | ||
+ | |||
+ | :<math>\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.</math> | ||
+ | |||
+ | The mean value theorem implies that if ''x'' > ''m'', then ''g''(''x'') ≠ ''g''(''m'') (since otherwise there would be a ''y'' in the interval (''m'', ''x'') with <math>g'(y)=0</math>). [[Cauchy's mean value theorem]] applied to the interval [''m'', ''x''] now implies that | ||
+ | |||
+ | :<math>\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.</math> | ||
+ | |||
+ | Since ''f'' converges to positive infinity, if ''x'' is large enough, then ''f''(''x'') ≠ ''f''(''m''). Write | ||
+ | |||
+ | :<math>\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.</math> | ||
+ | |||
+ | Now, | ||
+ | |||
+ | :<math> | ||
+ | \begin{align} | ||
+ | & \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ | ||
+ | & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ | ||
+ | & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. | ||
+ | \end{align} | ||
+ | </math> | ||
+ | |||
+ | For ''x'' sufficiently large, this is less than ''ε'' and therefore | ||
+ | |||
+ | :<math>\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.</math>* | ||
+ | * Note: Steps are missing. |
Latest revision as of 23:10, 2 January 2012
Contents
Discovered By
Named after Guillaume de l'Hopital; however, it is believed that the rule was actually discovered by Johann Bernoulli.
The Rule
For or
case, the limit
where
and
are the first derivatives of
and
, respectively.
Examples
Proof of l'Hôpital's rule
A standard proof of l'Hôpital's rule uses Cauchy's mean value theorem. l'Hôpital's rule has many variations depending on whether c and L are finite or infinite, whether f and g converge to zero or infinity, and whether the limits are one-sided or two-sided.
Zero over zero
Suppose that c and L are finite and f and g converge to zero.
First, define (or redefine) f(c) = 0 and g(c) = 0. This makes f and g continuous at c, but does not change the limit (since, by definition, the limit does not depend on the value at the point c).
Since exists, there is an interval (c − δ, c + δ) such that for all x in the interval, with the possible exception of x = c, both
and
exist and
is not zero.
If x is in the interval (c, c + δ), then the mean value theorem and Cauchy's mean value theorem both apply to the interval [c, x] (and a similar statement holds for x in the interval (c − δ, c)). The mean value theorem implies that g(x) is not zero (since otherwise there would be a y in the interval (c, x) with ). Cauchy's mean value theorem now implies that there is a point ξx in (c, x) such that
If x approaches c, then ξx approaches c (by the squeeze theorem ). Since exists, it follows that
Infinity over infinity
Suppose that L is finite, c is positive infinity, and f and g converge to positive infinity.
For every ε > 0, there is an m such that
The mean value theorem implies that if x > m, then g(x) ≠ g(m) (since otherwise there would be a y in the interval (m, x) with ). Cauchy's mean value theorem applied to the interval [m, x] now implies that
Since f converges to positive infinity, if x is large enough, then f(x) ≠ f(m). Write
Now,
- $\begin{align}
& \left|\frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - \frac{f(x)-f(m)}{g(x)-g(m)}\right| \\ & \quad \leq \left|\frac{f(x)-f(m)}{g(x)-g(m)}\right| \left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right| \\ & \quad < (|L|+\varepsilon)\left|\frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)} - 1\right|. \end{align}$ (Error compiling LaTeX. Unknown error_msg)
For x sufficiently large, this is less than ε and therefore
*
- Note: Steps are missing.