Math Doubts

L’Hopital’s Rule Proof

The l’hopital’s rule is a method of evaluating the limit of a rational expression when the limit of a rational expression is indeterminate as the input of the function approaches some value.

$\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f'(x)}{g'(x)}}$

In calculus, the l’hospital’s formula can be derived mathematically in three steps.

Limit of a Rational function

Let $f(x)$ and $g(x)$ be two functions in $x$. Assume, $c$ is a constant. The two functions formed a function in rational form. The limit of the quotient of $f(x)$ by $g(x)$ as $x$ approaches $c$ is written in the following mathematical form.

$\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$

Condition for Proving the Rule

If $f(x)$ and $g(x)$ are continuously differentiable at a real number $c$, and $f(c) \,=\, g(c) \,=\, 0$, then $f(x) \,=\, f(x)-f(c)$ and $g(x) \,=\, g(x)-g(c)$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)-f(c)}{g(x)-g(c)}}$

The limit of the rational expression equals to indeterminate as $x$ approaches $c$.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\dfrac{f(c)-f(c)}{g(c)-g(c)}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\dfrac{0}{0}$

It clears that the condition, which we have assumed above, makes the the limit of the rational expression is equal to indeterminate mathematically as the input of the function approaches a value.

Limit of function in Differentiation form

Now, comeback to the following mathematical equation to prove the l’Hopital’s rule.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)-f(c)}{g(x)-g(c)}}$

Each expression in the rational function must have an expression $x-c$ as its denominator for expressing each expression in rational function and it will be useful to us for obtaining it in differentiation form. So, let’s do it in mathematically acceptable manner.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \Bigg(\dfrac{f(x)-f(c)}{g(x)-g(c)} \times 1\Bigg)}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \Bigg(\dfrac{f(x)-f(c)}{g(x)-g(c)} \times \dfrac{x-c}{x-c}\Bigg)}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{\Big(f(x)-f(c)\Big)(x-c)}{\Big(g(x)-g(c)\Big)(x-c)}}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{\Big(f(x)-f(c)\Big)(x-c)}{\Big(g(x)-g(c)\Big)(x-c)}}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)-f(c)}{\dfrac{\Big(g(x)-g(c)\Big)(x-c)}{x-c}}}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)-f(c)}{(x-c) \times \dfrac{\Big(g(x)-g(c)\Big)}{x-c}}}$

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{\dfrac{f(x)-f(c)}{x-c}}{\dfrac{g(x)-g(c)}{x-c}}}$

Now, use the quotient rule of limits for bringing the mathematical proof to next level.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\dfrac{\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)-f(c)}{x-c}}}{\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{g(x)-g(c)}{x-c}}}$

The expressions $f(x)-f(c)$ and $g(x)-g(c)$ express the change in function $f(x)$ and the change in function $g(x)$ respectively. Similarly, the expression $x-c$ also expresses the chance in variable $x$.

According to the fundamental definition of the derivative, the quotient of $f(x)-f(c)$ by $x-c$ is called the derivative of function $f(x)$. In the same way, the quotient of $g(x)-g(c)$ by $x-c$ is called the derivative of function $g(x)$

The $x\,\to\,c$ means the value of $x$ is closer to $c$. So, the change in values of functions and change in variable are infinitesimal. Therefore, the derivative of both functions are expressed as follows.

$(1).\,\,\,$ $\dfrac{df(x)}{dx}$ $\,=\,$ $\dfrac{f(x)-f(c)}{x-c}$

$(2).\,\,\,$ $\dfrac{dg(x)}{dx}$ $\,=\,$ $\dfrac{g(x)-g(c)}{x-c}$

Now, we can replace the expressions in numerator and denominator by their respective derivatives of the functions.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\dfrac{\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{d}{dx}f(x)}}{\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{d}{dx}g(x)}}$

In differential calculus, the expressions $\dfrac{df(x)}{dx}$ and $\dfrac{dg(x)}{dx}$ are simply denoted by $f'(x)$ and $g'(x)$ respectively.

$\implies$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\dfrac{\displaystyle \large \lim_{x\,\to\,c}{\normalsize f'(x)}}{\displaystyle \large \lim_{x\,\to\,c}{\normalsize g'(x)}}$

Now, use the quotient rule of limits for simplifying the right hand side expression of the equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f(x)}{g(x)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize \dfrac{f'(x)}{g'(x)}}$

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved