Math Doubts

Proof of L’Hôpital’s Rule or L’Hospital’s Rule

The L’Hospital’s rule is expressed mathematically in the following form in calculus.

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

It is used as a formula to evaluate the limit of a rational function when the limit is indeterminate as the input approaches a value. Let us learn how to prove the l’hospital’s formula mathematically in calculus.

Limit of a function in Indeterminate form

Let $f(x)$ and $g(x)$ be two functions in terms of $x$, and assume that their limits are zero as the value of $x$ approaches a value $c$.

$(1).\,\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize f(x)}$ $\,=\,$ $f(c)$ $\,=\,$ $0$

$(2).\,\,$ $\displaystyle \large \lim_{x\,\to\,c}{\normalsize g(x)}$ $\,=\,$ $g(c)$ $\,=\,$ $0$

The function $f(x)$ divided by another function $g(x)$ forms a rational function and the limit of rational function is indeterminate as the value of $x$ tends to $c$.

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

An adjustment for deriving the L’Hôpital’s Rule

There is no value for the zero in mathematics. So, it can be subtracted from any function due to lack of change in the result.

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

We have assumed that the limits of functions $f(x)$ and $g(x)$ are zero as the value of $x$ is closer to $c$. Therefore, the zeros can be replaced by $f(c)$ and $g(c)$ in the expressions of rational function.

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

Express the function in Differentiation form

The expressions in both numerator and denominator of rational function are in terms of $x$ and $c$, and the terms in each expression are in subtraction form. So, let us try to include the literals $x$ and $c$ in the function in subtraction form.

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

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

Now, let us try to express the multiplication of two rational functions in derivative form by the multiplication rule of the fractions.

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

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

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

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

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

Write the second factor in reciprocal form as per the reciprocal rule of the fractions.

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

Now, multiply the fractions to find the product of them as per the multiplication of the fractions.

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

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

Observe the each expression in the rational function and every expression is in difference form. So, each expression in the rational function can be written in the form of derivatives.

$(1).\,\,$ $df(x) \,=\, f(x)-f(c)$

$(2).\,\,$ $dg(x) \,=\, g(x)-g(c)$

$(3).\,\,$ $dx \,=\, x-c$

Now, substitute them in the mathematical expression to express the rational function in differential form.

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

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

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

The derivatives of the functions $f(x)$ and $g(x)$ are denoted by $f'(x)$ and $g'(x)$ respectively in simple form in the differential calculus.

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

It is derived that the limit of a rational function when its limit is indeterminate, can be evaluated by the limit of rational function after differentiating the expressions in the numerator and denominator of the rational function. It is called the l’hôpital’s rule or l’hospital’s rule, and also called the bernoulli’s rule.

If in case, the limit is indeterminate,

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

Then, repeat the same procedure to express the l’hospital’s rule in the following form.

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

The limit of a rational function when the limit is indeterminate, can be obtained by finding the limit of rational expression after differentiating the numerator and denominator one more time. Repeat the same procedure until we get the limit of a rational function.

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