Let \( f, g\) be two functions defined on an interval \( I = [a,b]\) and their respective derivative \( f', g'\) defined on \(]a , b [\).
Now, let \( \alpha \in \overset{-}{I}\cup{\{\infty\}} \) be a point or an extremity of \( I\); \( \alpha \) be a point or an extremity, it can be whatever value, including \( -\infty \) or \( +\infty \).
In order to calculate the limit in \( \alpha\) of a function under the form of a quotient presenting an indeterminate form of type \( \left[ \frac{0}{0} \right]\) or \( \left[ \frac{\pm \infty}{\pm \infty} \right]\), there may be an interest in using this method, with the aim of removing indeterminacy.
L'Hôpital's rule tells us that:
$$ \forall x \in I, \enspace \forall \alpha \in \overset{-}{I}\cup{\{\infty\}}, \enspace g(x) \neq 0, \hspace{0.2em} g'(x) \neq 0, $$
$$ \left \{ \begin{align*} lim_{x \to \alpha} \ f(x) = 0, \\ lim_{x \to \alpha} \ g(x) = 0 \\ \\ lim_{x \to \alpha} \ \frac{f(x)}{g(x)} = l \end{align*} \right \} \Longrightarrow lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = l \qquad (L'H \it{ô} pital ) $$
$$ \left \{ \begin{align*} lim_{x \to \alpha} \ f(x) = \pm \infty, \\ lim_{x \to \alpha} \ g(x) = \pm \infty \\ \\ lim_{x \to \alpha} \ \frac{f(x)}{g(x)} = l \end{align*} \right \} \Longrightarrow lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = l \qquad (L'H \it{ô} pital^*) $$
Let \( f, g\) be two functions defined on \( I = [a,b]\) and \( \alpha\) a point or an extremity of \( I\), and their respective derivative \( f', g'\) which are non-zero at the same time, and the following hypothesis \( (H) \):
$$ \Biggl \{ \begin{align*} lim_{x \to \alpha} \ f(\alpha) = 0 \\ lim_{x \to \alpha} \ g(\alpha) = 0 \end{align*} \qquad (H) $$
If we want to calculate:
$$ lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} $$
We end up with an indeterminate form of type \( \left[ \frac{0}{0} \right]\):
$$ lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} = \left[ \frac{0}{0} \right] $$
Thanks to Cauchy's mean value theorem, we know that:
$$ \forall (f, g) \ continuous \ on \ [a,b]^2 \ and \ derivables \ sur \ ]a,b[^2, $$
$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} $$
In our case,
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \enspace \frac{f'(c)}{g'(c)} = \frac{f(x) - f(\alpha)}{g(x) - g(\alpha)} \qquad (1) $$
But, with \( (H) \) the expression \( (1) \) becomes \( (2) \):
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \enspace \frac{f'(c)}{g'(c)} = \frac{f(x)}{g(x)} \qquad (2) $$
Now, applying the limit when \( x \to \alpha \):
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \ lim_{x \to \alpha} \enspace \frac{f'(c)}{g'(c)} = lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} = l \qquad (3)$$
Now, as \( \alpha < c < x\), we do have these two implications:
$$ x \to \alpha \Longrightarrow c \to \alpha \Longrightarrow c \to x $$
Therefore, if \( x \to \alpha \), then \( c \to x \) and we can rewrite \( (3) \) as \( (3') \) :
$$ lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} = l \qquad (3') $$
And as a result,
$$ \forall x \in I, \enspace \forall \alpha \in \overset{-}{I}\cup{\{\infty\}}, \enspace g(x) \neq 0, \hspace{0.2em} g'(x) \neq 0, $$
$$ \left \{ \begin{align*} lim_{x \to \alpha} \ f(x) = 0, \\ lim_{x \to \alpha} \ g(x) = 0 \\ \\ lim_{x \to \alpha} \ \frac{f(x)}{g(x)} = l \end{align*} \right \} \Longrightarrow lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = l \qquad (L'H \it{ô} pital) $$
According to this rule, if: \( lim_{x \to \alpha} \enspace f'(x) \neq 0 \) or even \( lim_{x \to \alpha} \enspace g'(x) \neq 0 \), then, the limit of the quotient \( \frac{f}{g} \) can be easily obtained.
While we are still with an indeterminate form, we have to start the process again until the indeterminacy is removed.
In the same way, it can be shown that it also works when instead of \( (H)\) we do have a new hypothesis \( (H')\), such as:
$$ \Biggl \{ \begin{align*} lim_{x \to \alpha} \enspace f(x) = \pm \infty \\ lim_{x \to \alpha} \enspace g(x) = \pm \infty \end{align*} \qquad (H') $$
Applying the limit, we then again have an indeterminate form:
$$ lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} = \left[ \frac{\pm \infty}{\pm \infty} \right] $$
Let us rewrite \( (1) \) previously found:
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \enspace \frac{f'(c)}{g'(c)} = \frac{f(x) - f(\alpha)}{g(x) - g(\alpha)} \qquad (1) $$
Now, applying the limit when \( x \to \alpha \):
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \ lim_{x \to \alpha} \enspace \frac{f'(c)}{g'(c)} = lim_{x \to \alpha} \enspace \frac{f(x) - f(\alpha)}{g(x) - g(\alpha)}= l \qquad (4)$$
As \( f(\alpha) \) and \( g(\alpha) \) are both constants and knowing \( (H') \), we can write that:
$$ lim_{x \to \alpha} \enspace \frac{f(x) - f(\alpha)}{g(x) - g(\alpha)} = lim_{x \to \alpha} \ \frac{f(x)}{g(x)} = l \qquad (4') $$
So, thanks to \( (4) \) and \( (4') \) mixed together, we obtain \( (5)\):
$$ \forall x \in I, \enspace \exists c \in \hspace{0.05em} ]\alpha, x[, \ lim_{x \to \alpha} \enspace \frac{f'(c)}{g'(c)} = lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)}= l \qquad (5)$$
Finally, in the same as above, we do have a chain reaction when \(x \to a\), such as:
$$ x \to \alpha \Longrightarrow c \to \alpha \Longrightarrow c \to x $$
Therefore, if \( x \to \alpha \), then \( c \to x \) and we can rewrite \( (5) \) as \( (5') \) :
$$ lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = lim_{x \to \alpha} \enspace \frac{f(x)}{g(x)} = l \qquad (5') $$
And as a result,
$$ \forall x \in I, \enspace \forall \alpha \in \overset{-}{I}\cup{\{\infty\}}, \enspace g(x) \neq 0, \hspace{0.2em} g'(x) \neq 0, $$
$$\left \{ \begin{align*} lim_{x \to \alpha} \ f(x) = \pm \infty, \\ lim_{x \to \alpha} \ g(x) = \pm \infty \\ \\ lim_{x \to \alpha} \ \frac{f(x)}{g(x)} = l \end{align*} \right \} \Longrightarrow lim_{x \to \alpha} \enspace \frac{f'(x)}{g'(x)} = l \qquad (L'H \it{ô} pital^*)$$
At last, with a product \( fg \) having an indeterminate form of type \( [0] \times {[ \pm \infty]}\) applying the limit on \( \alpha\), such as:
$$ lim_{x \to \alpha} \enspace f(x) g(x) = \bigl[ 0 \times \pm \infty \bigr]$$
$$ with \enspace \Biggl \{ \begin{align*} lim_{x \to \alpha} \enspace f(x) = 0 \\ lim_{x \to \alpha} \enspace g(x) = \pm \infty \end{align*} $$
We can rather considerate the product as a quotient, to be able to apply the rule, so that:
$$ lim_{x \to \alpha} \enspace f(x) g(x) = lim_{x \to \alpha} \enspace \frac{f(x)}{\frac{1}{g(x)}}$$
$$ lim_{x \to 0^+} \enspace \frac{sin(x)}{x} = \left[ \frac{0^+}{0^+} \right] $$
We are facing an indeterminate form of type \( \left[ \frac{0}{0} \right]\).
Applying the rule, we do have:
$$ lim_{x \to 0^+} \enspace \frac{sin(x)}{x} = lim_{x \to 0} \enspace \frac{cos(x)}{1}= 1 $$
$$ lim_{x \to +\infty} \enspace \frac{e^x}{x^n} = \left[ \frac{+ \infty}{+ \infty} \right] $$
Applying the rule in chain, the numerator will remain fixed, but the denominator will become a constant over the \(n\)-th derivation:
$$ lim_{x \to +\infty} \enspace \frac{e^x}{x^n} = lim_{x \to +\infty} \enspace \frac{e^x}{nx^{n-1}} $$
$$ lim_{x \to +\infty} \enspace \frac{e^x}{x^n} = lim_{x \to +\infty} \enspace \frac{e^x}{n(n-1)x^{n-2}} $$
$$ lim_{x \to +\infty} \enspace \frac{e^x}{x^n} \enspace = \enspace... \enspace = \enspace lim_{x \to +\infty} \enspace \frac{e^x}{n!} = + \infty $$
$$ lim_{x \to 0^+} \ x^n ln(x) = \bigl[ 0^+ \times - \infty \bigr] $$
$$ with \enspace \Biggl \{ \begin{align*} lim_{x \to 0^+} \ x^n = \ 0^+ \\ lim_{x \to 0^+} \ ln(x) = - \infty \end{align*} $$
To apply the rule, let us considerate a quotient to calculate the limit:
$$ lim_{x \to 0^+} \ x^n ln(x) = lim_{x \to 0^+} \ \frac{ln(x)}{x^{-n}} $$
At this stage, applying the rule we do have:
$$ lim_{x \to 0^+} \ x^n ln(x) = lim_{x \to 0^+} \ \frac{1 \over x}{-nx^{-n-1}} $$
$$ lim_{x \to 0^+} \ x^n ln(x) = lim_{x \to 0^+} \ \frac{1}{-nx^{-n}} $$
$$ lim_{x \to 0^+} \ x^n ln(x) = lim_{x \to 0^+} \ - \frac{x^{n}}{n} = - \infty $$
$$ lim_{x \to 0^+} \ \frac{ln(1 + x)}{x} = \left[ \frac{0^+}{0^+} \right] $$
$$ lim_{x \to 0^+} \ \frac{ln(1 + x)}{x} = lim_{x \to 0^+} \ \frac{\frac{1}{1 + x}}{1} = 1 $$
$$ lim_{x \to 0^+} \ \frac{e^x - 1}{x} = \left[ \frac{0^+}{0^+} \right] $$
$$ lim_{x \to 0^+} \ \frac{e^x - 1}{x} = lim_{x \to 0^+} \ \frac{e^x}{1} = 1 $$