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.

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

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