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L'Hospital's Rule

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L'Hospital's Rule

If f(x)=fn(x)fd(x)$f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)}}$, where

f(x)x=a=00$f \left(x\right) {|}_{x = a} = \frac{0}{0}$;

fn(x)x=a=0$\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a} = 0$ and

fd(x)x=a=0$\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a} = 0$, then

limxaf(x)${\lim}_{x \to a} f \left(x\right)$     =[ddxfn(x)]xa$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{\left[\frac{d}{\mathrm{dx}} {f}_{n} \left(x\right)\right] {|}_{x \to a}}$ $\quad \quad \quad \div \textcolor{c \mathmr{and} a l}{\left[\frac{d}{\mathrm{dx}} {f}_{d} \left(x\right)\right] {|}_{x \to a}}$
when the numerator and denominator are differentiable.

Note: The limit is the slope of numerator divided by slope of denominator at $x = a$.

introduction

The function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)}}$ such that $\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a} = 0$ and $\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a} = 0$. It was discussed that the slope of the numerator and denominator defines the limits. This is formally given by L'Hospital's Rule.

There are multiple proofs for L'Hospital's Rule. The discussion on slopes here is the intuitive understanding (not a formal proof) of L'Hospital's Rule.

formal handling with slop

Given the function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)}}$ such that $\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a} = 0$ and $\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a} = 0$.

$f \left(x\right) {|}_{x = a + \delta}$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a + \delta} \div \textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a + \delta}$
$\quad \quad = \left[\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a + \delta} - {f}_{n} \left(a\right)\right]$
$\quad \quad \quad \quad \div \left[\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a + \delta} - {f}_{d} \left(a\right)\right]$
as ${f}_{n} \left(a\right) = 0$ and ${f}_{d} \left(a\right) = 0$.

$\quad \quad = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a + \delta} - {f}_{n} \left(a\right)}{\delta}$
$\quad \quad \quad \quad \div \frac{\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a + \delta} - {f}_{d} \left(a\right)}{\delta}$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{\textrm{s l o p e} {f}_{n} \left(x\right) {|}_{x = a}} \div \textcolor{c \mathmr{and} a l}{\textrm{s l o p e} {f}_{d} \left(x\right) {|}_{x = a}}$.

The beauty of this mathematical manupulation is that the function evaluated at a position is $\frac{0}{0}$ which is not deterimined to be a value. The same is equivalently given as slope of numerator divided by slope of dinominator which provides another way to determine the value.

revisit

If you have started on the calculus and limits, then you may not have come across derivative, differentiation, and differentiability. If required, you may have to revisit this page when you have completed the differential calculus.

For the limit of a function, evaluate the function formed by derivatives of the numerator and the denominator.

example

Given function $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{x}^{2} - 1}}{\textcolor{c \mathmr{and} a l}{x - 1}}$. what is ${\lim}_{x \to 1} f \left(x\right)$?

The answer is '$2$'.

Differentiating numerator $\frac{d}{\mathrm{dx}} \textcolor{\mathrm{de} e p s k y b l u e}{\left({x}^{2} - 1\right) {|}_{x = 1} = 2}$
Differentiating denominator $\frac{d}{\mathrm{dx}} \textcolor{c \mathmr{and} a l}{\left(x - 1\right) {|}_{x = 1} = 1}$

${\lim}_{x \to 1} f \left(x\right)$
$\quad \quad = \frac{2}{1}$

summary

L'Hospital's Rule: If $f \left(x\right) = \frac{\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)}}{\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)}}$, where
$f \left(x\right) {|}_{x = a} = \frac{0}{0}$;
$\textcolor{\mathrm{de} e p s k y b l u e}{{f}_{n} \left(x\right)} {|}_{x = a} = 0$ and
$\textcolor{c \mathmr{and} a l}{{f}_{d} \left(x\right)} {|}_{x = a} = 0$, then
${\lim}_{x \to a} f \left(x\right)$
$\quad \quad = \textcolor{\mathrm{de} e p s k y b l u e}{\left[\frac{d}{\mathrm{dx}} {f}_{n} \left(x\right)\right] {|}_{x \to a}}$
$\quad \quad \quad \quad \div \textcolor{c \mathmr{and} a l}{\left[\frac{d}{\mathrm{dx}} {f}_{d} \left(x\right)\right] {|}_{x \to a}}$
when the numerator and denominator are differentiable.

Outline