L'Hôpital's rule

format_list_bulleted Contenido keyboard_arrow_down

ImprimirCitar

Guillaume de l'Hôpital, was the one who announced this rule.

In mathematics, more specifically in differential calculus, the l'Hôpital rule or l'Hôpital-Bernoulli rule is a rule that use derivatives to help evaluate limits of functions that are in indeterminate form.

This rule is named after the 17th-century French mathematician Guillaume François Antoine, Marquis de l'Hôpital (1661 - 1704), who introduced the rule in his work Analyse des infiniment petits pour l'intelligence des lignes courbes (1696), the first text ever written on differential calculus, although the rule is now known to be due to Johann Bernoulli, who developed and demonstrated it. The explanation is that both had entered into a curious business arrangement through which the Marquis de L'Hopital bought the rights to Bernoulli's mathematical discoveries.

Statement

La rule of L'Hôpital is a consequence of the Cauchy medium value theorem that is given only in the case of indeterminations of the type $frac{0}{0}$ or $frac{infty}{infty}$ .

Sean. f and g two continuous functions defined in the interval [a,b], derivable in (a,b) and be it c belonging to (a,b) such that f(c♪g(c♪0 and g'(xIt was xI was.c.
If there is a limit L of f '/g' in c, then there is the limit of f/g (in c) and is equal to L. So,
$lim _{{xto c}}{f(x) over g(x)}=lim _{{xto c}}{f'(x) over g'(x)}=L$

Guillaume de l'Hôpital

Demo

The following argument can be taken as a "demonstration" of the L'Hôpital rule, although in fact a rigorous demonstration requires type arguments $varepsilon$ - $delta$ more delicate.

Like ${displaystyle g(c)=0}$ and ${displaystyle g'(x)neq 0}$ Yeah. ${displaystyle xneq c}$ You have to. ${displaystyle g(x)neq 0}$ Yeah. ${displaystyle xneq c}$ as a result of the Rolle Theorem.

Given that f(c♪g(c)=0, applying the Medium Value Theorem of Cauchy, for everything x in (a,b), with x other than cexists t_x in the end interval a and bsuch that the quotient f(x)/g(x) can be written as follows:

${cfrac {f(x)}{g(x)}}={cfrac {f(x)-f(c)}{g(x)-g(c)}}={cfrac {f'(t_{x})}{g'(t_{x})}}$

When x towards cequating the values of the equalitys above, t_x also tends towards cSo...

${displaystyle lim _{xto c}{cfrac {f(x)}{g(x)}}=lim _{xto c}{cfrac {f'(t_{x})}{g'(t_{x})}}=L}$

Note: the last step to the limit, while true, would require more rigorous justification.

Examples

L'Hôpital's rule is applied to save indeterminacies that result from replacing the numerical value by taking the given functions to the limit. The rule says to derive the numerator and denominator separately; that is: let the original functions be f(x)/g(x), when applying the rule you will get: f'(x)/g'(x).

Simple application

{displaystyle lim _{xto 0}{frac {x-sin(x)}{x^{3}}}={cfrac {0}{0}}}

{displaystyle lim _{xto 0}{frac {sin(x)}{x}}quad {xrightarrow {mathrm {l'H{hat {o}}pital} }}quad lim _{xto 0}{frac {cos(x)}{1}}}

{displaystyle ={frac {1}{1}}=1}

Consecutive application

As long as the function is n times continuous and differentiable, the rule can be applied n times:

lim_{x to 0} frac{e^x-e^{-x}-2x}{x-sin(x)}

xrightarrow{mathrm{l'H hat{o} pital}} quad lim_{x to 0} frac{e^x-(-e^{-x})-2}{1-cos(x)}

xrightarrow{mathrm{l'H hat{o} pital}} quad lim_{x to 0} frac{e^x-e^{-x}}{sin(x)}

xrightarrow{mathrm{l'H hat{o} pital}} quad lim_{x to 0} frac{e^x-(-e^{-x})}{cos(x)} = frac{e^0+e^{-0}}{cos(0)} = frac{1+1}{1} = 2

Algebraic adaptations

Given the usefulness of the rule, it is practical to transform other types of indeterminations to the type $begin{matrix} frac{0}{0} end{matrix}$ by algebraic transformations:

Incompatible quotients

Type indeterminations $begin{matrix} frac{infty}{infty} end{matrix}$ can be transformed by double-investment of quotients:

lim_{x to infty} cfrac{x^4}{x} = lim_{x to infty} cfrac{cfrac{1}{x}}{cfrac{1}{x^4}}

This way you can prove that type indeterminations $begin{matrix} frac{infty}{infty} end{matrix}$ They can also be resolved through the application of the L'Hôpital rule directly, without the application of double investment.

Non-quotient indeterminacies

Sometimes some undetermined limits that are not presented as quotients can be resolved with this rule, using previous transformations that lead to a quotient of the type ${displaystyle {cfrac {0}{0}}}$ or ${displaystyle {cfrac {infty }{infty }}}$ .

Type ${displaystyle 0cdot infty }$

It's about making a transformation like

{displaystyle 0cdot infty ={cfrac {0}{cfrac {1}{infty }}}={cfrac {0}{0}}}

{displaystyle 0cdot infty ={cfrac {infty }{cfrac {1}{0}}}={cfrac {infty }{infty }}}

The most classic:

{displaystyle lim _{xto 0}xcdot log(x)=lim _{xto 0}{cfrac {log(x)}{cfrac {1}{x}}}quad {xrightarrow {mathrm {l'H{hat {o}}pital} }}quad lim _{xto 0}{cfrac {cfrac {1}{x}}{cfrac {-1}{x^{2}}}}=lim _{xto 0}(-x)=0}

Type ${displaystyle infty -infty }$

lim_{x to infty} x - sqrt{x^2 - x} =

= lim_{x to infty} cfrac{left(x + sqrt{x^2 - x}right)left(x - sqrt{x^2 - x}right)}{x + sqrt{x^2 - x}} = lim_{x to infty} cfrac{x^2 - (x^2 - x)}{x + sqrt{x^2 - x}} = lim_{x to infty} cfrac{x}{x + sqrt{x^2 - x}}

= lim_{x to infty} cfrac{ overbrace{x}^{infty} } { underbrace{x + sqrt{x^2 - x}}_{infty} } = cfrac{infty}{infty}

lim_{x to infty} cfrac{x}{x + sqrt{x^2 - x}} quad xrightarrow{mathrm{l'H hat{o} pital}} quad lim_{x to infty} cfrac{(x)'}{(x + sqrt{x^2 - x})'} =

= lim_{x to infty} cfrac{1}{1 + cfrac{2x - 1}{2 sqrt{x^2 - x}}} = cfrac{1}{1 + 1} = cfrac{1}{2}

Generalizations

The rule of L'Hôpital is valid for lateral limits, limits in infinite and infinite limits.

The L'Hôpital rule can be extended to scaling functions n variables that are different. Two different functions f and g such that f(c) = g(c) = 0, you have:

$lim_{mathbf{x}to mathbf{c}} frac{f(mathbf{x})}{g(mathbf{x})} = lim_{mathbf{x}to mathbf{c}} frac{nabla f(mathbf{x})cdot(mathbf{x-c})}{nabla g(mathbf{x})cdot(mathbf{x-c})}= lim_{mathbf{x}to mathbf{c}} frac{|nabla f(mathbf{x})|}{|nabla g(mathbf{x})|} left(frac{cos theta_f}{cos theta_g} right)$