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

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==Proof of l'Hôpital's rule==
 
==Proof of l'Hôpital's rule==
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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. All the variations follow from the two main variations below without, for the most part, requiring any new reasoning.<ref>{{Cite book | last=Spivak | first=Michael | authorlink=Michael Spivak | title=Calculus | year=1994 | publisher=Publish or Perish | location=Houston, Texas | isbn=0-914098-89-6 | pages=201–202, 210–211}}</ref>
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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.  
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===Zero over zero===
 
===Zero over zero===
 
Suppose that ''c'' and ''L'' are finite and ''f'' and ''g'' converge to zero.
 
Suppose that ''c'' and ''L'' are finite and ''f'' and ''g'' converge to zero.

Revision as of 21:59, 15 March 2011

Discovered By

Guillaume de l'Hopital

The Rule

For $\frac {0}{0}$ or $\frac {\infty}{\infty}$ case, the limit \[\lim_{x \to a} \cfrac {f(x)}{g(x)} = \lim_{x \to a} \cfrac {f'(x)}{g'(x)}\] where $f'(x)$ and $g'(x)$ are the first derivatives of $f(x)$ and $g(x)$, respectively.

Examples

$\lim_{x \to 4} \cfrac {x^3 - 64}{4 - x} = \lim_{x \to 4} \cfrac {3x^2}{-1} = -3(4)^2 = -3(16) = \boxed {-48}$

$\lim_{x \to 0} \cfrac {\sin x}{x} = \lim_{x \to 0} \cfrac {\cos x}{1} = \cos (0) = \boxed {1}$

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 $\lim_{x\to c}f'(x)/g'(x)$ 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 $f'(x)$ and $g'(x)$ exist and $g'(x)$ 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 $g'(y)=0$). Cauchy's mean value theorem now implies that there is a point ξx in (c, x) such that

$\frac{f(x)}{g(x)} = \frac{f'(\xi_x)}{g'(\xi_x)}.$

If x approaches c, then ξx approaches c (by the squeeze theorem ). Since $\lim_{x\to c}f'(x)/g'(x)$ exists, it follows that

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(\xi_x)}{g'(\xi_x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}.$

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

$\left|\frac{f'(x)}{g'(x)} - L\right| < \varepsilon \quad \text{for } x\geq m.$

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 $g'(y)=0$). Cauchy's mean value theorem applied to the interval [m, x] now implies that

$\left|\frac{f(x)-f(m)}{g(x)-g(m)} - L\right| < \varepsilon \quad \text{for } x>m.$

Since f converges to positive infinity, if x is large enough, then f(x) ≠ f(m). Write

$\frac{f(x)}{g(x)} = \frac{f(x)-f(m)}{g(x)-g(m)} \cdot \frac{f(x)}{f(x)-f(m)} \cdot \frac{g(x)-g(m)}{g(x)}.$

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

$\left|\frac{f(x)}{g(x)} - L\right| < 2\varepsilon.$*
  • Note: Steps are missing.
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