# 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 :

1. 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?
2. 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 $$x, then $$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 ?

• The highlighted lines in the proof need a detailed analysis. Obviously "this is not obviously possible" but rather involves some tricky argument. One such argument is available in my blog post : paramanands.blogspot.com/2013/11/… May 6, 2019 at 9:28
• @ParamanandSingh Thanks! Would you mind considering my second question ? May 6, 2019 at 9:37
• The variable $y$ remains fixed in this approach. It is not supposed to tend to $a$. I wonder what is the intent of the author here. May 6, 2019 at 9:40
• @ParamanandSingh I tried to remedy the author(Zorichs)’s proof, and finally got an understandable proof. Would you mind checking it out? math.stackexchange.com/a/3217921/34603 May 8, 2019 at 3:08

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.

• Minor typo: since $x_0$ can be chosen... " you need to replace this with $x$. Apart from that one needs to use $\epsilon, \delta$ stuff. For example when $x\to a^{+}$ the variable $c$ also changes. You can't say that the Right hand side tends to $f'(c) /g'(c)$. Another option is to use $\liminf$ and $\limsup$. May 8, 2019 at 3:14
• @ParamanandSingh Thanks ! Yes, the above is not a formal mathematical proof, but rather my reasoning process . (1) I don’t know what you mean by ‘you need to replace this with $x$’, the author’s $\lbrack x,y\rbrack$ is replaced by $\lbrack x,x_{0}\rbrack$ in my proof. .. May 8, 2019 at 3:43
• @ParamanandSingh ...(2) I admit that it is not easy to figure out the tendency of $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$ as $x \rightarrow a^{+}$, so I choose to leave it aside there and let’s see what the right hand side tends to as $x \rightarrow a^{+}$, as I’ve said, it tends to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$, that’s why I say the right hand side tends to $\frac{f^{'}\left( c \right)}{g^{'}\left( c \right)}$. May 8, 2019 at 3:43
• You need to write "since $x$ can be chosen arbitrarily in $(a, x_0)$, we can choose an $x$ near enough to the right side of $a$". May 8, 2019 at 4:09
• You can't leave $f'(c) /g'(c)$ like that. You need to replace it with limit. But unfortunately as $x\to a^+$ the variable $c$ does not necessarily tend to $a$ and hence one does not know its limiting behavior. I will post an answer based on $\limsup, \limsup$ which sort of fixes all this. May 8, 2019 at 4:12

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.

• So is your meaning equivalent to mine in the proof ? math.stackexchange.com/a/3217921/34603 May 8, 2019 at 3:08
• My post was not a proof but only a sketch. Anyway there is a similar but easier-to-follow proof in Pugh’s mathematical analysis textbook
– user555729
May 8, 2019 at 8:55

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 $$a and $$\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 $$x_n. Then $$x_n\to a$$ guarantees $$N(n)\to +\infty$$. Let $$y_n=z_{N(n))}$$. Then clearly $$a and $$y_n\to$$a\$.

Thus in both cases we obtain a sequence $$y_n\to a^+$$ such that $$a for all $$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.$$

• I think your proof is somehow what the author meant.
– user555729
May 9, 2019 at 20:07

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 $$L-\epsilon<\frac{f'(x)} {g'(x)} whenever $$a. 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 $$a and some $$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 $$a. Let $$\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 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 $$L=-\infty$$.

• You significantly changed the proof of the author
– user555729
May 8, 2019 at 8:49
• @User12239: author's proof was faulty/incomplete and I have provided one way to fix his approach. Another approach is in my blog post (see my comments to original question). May 8, 2019 at 14:58