Zorich's proof on L’Hospital’s Rule The following proof on L’Hospital’s Rule is excerpted from Zorich's Mathematical Analysis I, I have questions concerning the selected part in the proof :


*

*Why it is obviously possible to have $\dfrac {f\left( y\right) }{g\left( x\right) }\rightarrow 0$ and $\dfrac {g\left( y\right) }{g\left( x\right) }\rightarrow 0$ as $x\rightarrow a^{+}$ and $y\rightarrow a^{+}$, especially under the  hypothesis $2^0$ ? Anyone can explain it in detail?

*The proof also requires $y\rightarrow a^{+}$, so I found the
conclusion the proof come to is NOT $\dfrac {f\left( x\right)
}{g\left( x\right) }\rightarrow A$  as $x\rightarrow a^{+}$, but
should also added with the condition $y\rightarrow a^{+}$, and since
$x<y$, then $x\rightarrow a^{+}$ could be ensured by $y\rightarrow
a^{+}$, so I think the proof only proved that $\dfrac {f\left(
x\right) }{g\left( x\right) }\rightarrow A$  as $y\rightarrow a^{+}$
( but not $x\rightarrow a^{+}$). In a nutshell, I think the proof
didn't achieve the goal of proving L’Hospital’s Rule, am I all right
?





 A: The following is my remedy on the author(Zorichs)’s proof.
I replaced his $y$ with $x_{0}$, $\xi$ with $c$, then take the equality as
$$\begin{matrix}
\frac{f\left( x \right)}{g\left( x \right)} = \frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}\left( 1 - \frac{g\left( x_{0} \right)}{g\left( x \right)} \right) + \frac{f\left( x_{0} \right)}{g\left( x \right)} \\
\end{matrix}$$
Since $x_{0}$ can be chosen arbitrarily in $(a,\ x_{0})$, we can chose a $x_{0}$ near enough to the right side of $a$, then $\frac{f^{'}\left( x \right)}{g^{'}\left( x \right)}$ can be made as close as we please to $L$ in $(a,\ x_{0})$ due to $\lim_{x \rightarrow a^{+}}\frac{f^{'}\left( x \right)}{g^{'}\left( x \right)} = L$ and$\ x < x_{0}$, so can $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ for $c < x_{0}$. As $1 - \frac{g\left( x_{0} \right)}{g\left( x \right)} \rightarrow 1$ and $\frac{f\left( x_{0} \right)}{g\left( x \right)} \rightarrow 0$ for $\lim_{x \rightarrow a^{+}}{g\left( x \right)} = \infty$, the right side of the equality tends to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ as $x \rightarrow a^{+}$, which in turn is close to $L$, so $\frac{f\left( x \right)}{g\left( x \right)} \rightarrow L$ as $x \rightarrow a^{+}$, this is what the ∞/∞ case L’Hospital’s Rule seeks.

Update:
I will clarify what I mean by the right hand side tends to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ as $x \rightarrow a^{+}$ using the $\epsilon,\delta$ stuff.
The difference between $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ and the right hand side of the equality is : $|\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)} - \left\lbrack \frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}\left( 1 - \frac{g\left( x_{0} \right)}{g\left( x \right)} \right) + \frac{f\left( x_{0} \right)}{g\left( x \right)} \right\rbrack|$=$\ |\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)} \times \frac{g\left( x_{0} \right) - f\left( x_{0} \right)}{g\left( x \right)}|$.
Since $\lim_{x \rightarrow a}{g\left( x \right)} = \infty$，for every $\epsilon_{3} > 0$, there is a $\delta_{3} > 0$ such that $|\frac{g\left( x_{0} \right) - f\left( x_{0} \right)}{g\left( x \right)}| < \varepsilon_{3}$ where ${x\epsilon(a,a + \delta}_{3})$.
Since $\lim_{x \rightarrow a}\frac{f^{'}(x)}{g^{'}\left( x \right)} = L$, for every $\epsilon_{1} > 0$, there is a $\delta_{1} > 0$ such that $L - \epsilon_{1} < \frac{f^{'}\left( x \right)}{g^{'}\left( x \right)} < {L + \epsilon}_{1}$ where ${x\epsilon(a,a + \delta}_{1})$, then $|\frac{f^{'}\left( x \right)}{g^{'}\left( x \right)}| < {|L| + \epsilon}_{1}$. In order to ensure $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ also satisfy the inequality, one can choose $x_{0} = a + \delta_{1}$ so that $c$ of $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ within ${(a,a + \delta}_{1})$. Therefore, 
$$|\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)} \times \frac{g\left( x_{0} \right) - f\left( x_{0} \right)}{g\left( x \right)}| < |\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}| \times \varepsilon_{3} < ({\left| L \right| + \epsilon}_{1}) \times \varepsilon_{3}$$
Because $\varepsilon_{3}$ can be chosen arbitrarily small, which in turn can make $({\left| L \right| + \epsilon}_{1}) \times \varepsilon_{3}$ arbitrarily small as well, this is what I mean by the right hand side tends to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ as $x \rightarrow a^{+}$, it is equivalent to say $\frac{f\left( x \right)}{g\left( x \right)} = \frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}\left( 1 - \frac{g\left( x_{0} \right)}{g\left( x \right)} \right) + \frac{f\left( x_{0} \right)}{g\left( x \right)}$ is close to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ as $x \rightarrow a^{+}$, so both share the same limit.
A: As to your second question which you haven’t found a solution for, note that $\frac{f(x)}{g(x)}$ equals $\frac{f(y)}{g(x)}+\frac{f’(\zeta(y))}{g’(\zeta(y))}(1-\frac{g(y)}{g(x)})$ for every $y$, so you can make $y$ approach any quantity in the right side of this equality without affecting the left side.
A: Fix a sequence $x_n\to a^+$.


*

*Under $1^o$, note that for fixed $x>a$, we have $\frac{f(y)}{g(x)}\to 0$ and $\frac{g(y)}{g(x)}\to 0$ as $y\to a^+$. Thus for our given $x_n$, we can find $y_n$ such that $a<y_n<x_n$ and $\left|\frac{f(y_n)}{g(x_n)}\right|<\frac1n$ and $\left|\frac{g(y_n)}{g(x_n)}\right|<\frac1n$.

*Under condition $2^o$, for fixed $y>a$, we have $\frac{f(y)}{g(x)}\to 0$ and $\frac{g(y)}{g(x)}\to 0$ as $x\to a^+$. Let $z_n:=a+\frac{b-a}n$. Then there exists $\delta_n>0$ such that and $\left|\frac{f(z_n)}{g(x)}\right|<\frac1n$ and $\left|\frac{g(z_n)}{g(x)}\right|<\frac1n$ for all $x\in \left]a,a+\delta_n\right[$.
We may assume wlog that $\delta_n\to 0$.
Now for our given $x_n>a$ let $N(n)$ be maximal such that $x_n<a+\delta_{N(n)}$.
Then $x_n\to a$ guarantees $N(n)\to +\infty$. Let $y_n=z_{N(n))}$.
Then clearly $a<x_n<y_n$ and $y_n\to $a$.
Thus in both cases we obtain a sequence $y_n\to a^+$ such that $a<x_n< y_n$ for all $n$ and $\frac{f(y_n)}{g(x_n)}\to 0$ and $\frac{g(y_n)}{g(x_n)}\to 0$. With the accordingly found sequence $\xi_n$, we have $\xi_n\to a$ and hence
$$ \frac{f(x_n)}{g(x_n)}=\frac{f(y_n)}{g(x_n)}+\frac{f(\xi_n)}{g(\xi_n)}\left(1-\frac{g(y_n)}{g(x_n)}\right)\to A.$$
(This is true even if $A$ is infinite).
As $x_n$ was an arbitrary sequence with $x_n\to a$, we conclude 
$$ \frac{f(x)}{g(x)}\to A.$$
A: Here is one way to prove the second part of L'Hospital's Rule which deals with the form "$\text{anything} /\infty $". 
Let $\epsilon >0$ be arbitrary and then we choose a $\delta_1>0$ such that $$L-\epsilon<\frac{f'(x)} {g'(x)} <L+\epsilon\tag{1}$$ whenever $a<x<a+\delta_1$. Next choose a fixed number $x_0\in(a,a+\delta_1)$ and then we have the identity $$\frac{f(x)} {g(x)} =\frac{f'(c_x)} {g'(c_x)} \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}\tag{2}$$ for all $x$ with $a<x<x_0$ and some $c_x\in(x, x_0)$.
Since $g(x_0)/g(x)\to 0$ as $x\to a^+$ we can find another $\delta_2>0$ such that $$\left|\frac{g(x_0)}{g(x)}\right|<1\tag{3}$$ whenever $a<x<a+\delta_2$. Let $$\delta=\min(x_0-a,\delta_2)$$ and then we have the inequality $$(L-\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)} <\frac{f(x)} {g(x)} < (L+\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}$$ whenever $a<x<a+\delta$ using $(1),(2),(3)$.
Now taking limits as $x\to a^+$ we get $$L-\epsilon\leq\liminf_{x\to a^+} \frac{f(x)} {g(x)} \leq\limsup_{x\to a^+} \frac{f(x)} {g(x)} \leq L+\epsilon$$ Since $\epsilon $ is arbitrary it follows that $f(x) /g(x) \to L$ as $x\to a^+$.
Note that $x_0$ is fixed in above proof and the variable $c_x$ lying in $(x, x_0)$ does not necessarily tend to $a$ as $x\to a^+$. Hence we can't the predict the limiting behavior of ratio $f'(c_x) /g'(c_x) $. But we know that this ratio lies between $L-\epsilon $ and $L+\epsilon$ and that is sufficient to predict the limiting behavior of $f/g$. 
The same proof works with minor modifications if $L=\infty$ or $L=-\infty$. 
