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
$x&lt;y$, 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
?
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.
Here is one way to prove the second part of L'Hospital's Rule which deals with the form "$\text{anything} /\infty $". 

Let $\epsilon &gt;0$ be arbitrary and then we choose a $\delta_1&gt;0$ such that $$L-\epsilon&lt;\frac{f'(x)} {g'(x)} &lt;L+\epsilon\tag{1}$$ whenever $a&lt;x&lt;a+\delta_1$. 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 $a&lt;x&lt;x_0$ and some $c_x\in(x, x_0)$.

Since $g(x_0)/g(x)\to 0$ as $x\to a^+$ we can find another $\delta_2&gt;0$ such that $$\left|\frac{g(x_0)}{g(x)}\right|&lt;1\tag{3}$$ whenever $a&lt;x&lt;a+\delta_2$. Let $$\delta=\min(x_0-a,\delta_2)$$ and then we have the inequality $$(L-\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)} &lt;\frac{f(x)} {g(x)} &lt; (L+\epsilon) \left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}$$ whenever $a&lt;x&lt;a+\delta$ 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 $L=-\infty$.