# Mean Value Theorem, Indeterminate forms and L'Hospital's rule

I had my own doubts about the above equation. So I plotted a few graphs. I have taken $f(x) = x^2$ and $g(x) =e^x - 1$. I have chosen $a=-1$ and $b=1$. Below, are the two graphs I have plotted.

Here, $\frac{f(b) - f(a)}{b-a} =0$. Mean Value Theorem gives us $C_1=0$.

Above, $\frac{g(b) - g(a)}{b-a} =1.1751$. Mean Value Theorem gives us $C_2=0.1613$. Clearly, $C_1\neq C_2$. The equation in general can be written as, $$\frac{f'(C_1)}{g'(C_2)} = \frac{f(b) - f(a)}{g(b) - g(a)} \neq \frac{f'(C_1)}{g'(C_1)}(or) \frac{f'(C_2)}{g'(C_2)}$$

The proof of L'Hospital's rule was worked out using the equation $\frac{f'(C)}{g'(C)} = \frac{f(b) - f(a)}{g(b) - g(a)}$. It does not convince me to a good extent. Am I missing a point here, thereby complicating the work?

I went ahead and tried approaching the proof of L'Hospital's rule in a different way. If the below proof is right, then I'm quite sure that it is not a new way of approach.

Given, functions $f(x)$ and $g(x)$ are continuous and differentiable in an interval containing a point $a$. The two functions take zero values at $a$, say. We have, $\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$. This limit, at first look, takes an indeterminate form $\frac{0}{0}$. An indeterminate form can take any value( or no value ). We can write, $$\lim_{x\rightarrow a} \frac{f(x)h}{g(x)h}$$ Where $h\neq0$. Further we can write, $$\lim_{h\rightarrow 0} \frac{(f(a+h)-f(a-h))2h}{(g(a+h)-g(a-h))2h}$$ Where once again the limit takes the same indeterminate form. Mean Value Theorem gives us, $$\lim_{h\rightarrow 0} \frac{f'(C_1)}{g'(C_2)}$$

As $h$ gets close to zero, $C_1$ and $C_2$ get close to $a$. We can then write, $\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}= \lim_{x\rightarrow a} \frac{f(x)}{g(x)}$. Here, we assume that $g'(x) \neq 0$.

Could there be any flaw in this approach?

• You say you plotted $x^2$ and $e^x-1$, but then below is the plot of $x^2$ and $1$. Please clean up your question, it's very confusing. – 5xum Jul 31 '17 at 12:57
• The equations you see other than the functions are secants derived from the Mean Value Theorem. Did I clear it for you? – R004 Jul 31 '17 at 12:59
• It appears that you are doubting the truth of Cauchy's MVT. It is better that you see the proof. Also understand that it can't be derived using MVT separately applied on $f, g$. See this question math.stackexchange.com/q/1380881/72031 for more details. – Paramanand Singh Jul 31 '17 at 16:37
• The second part dealing with L'Hospital's Rule does not seem to make much sense. The only way to link the expression $f/g$ with $f'/g'$ is via Cauchy's MVT. Your approach does get to $f'/g'$ but the arguments of $f', g'$ are different and then it is not guaranteed that limit of $f'(c_1)/g'(c_2)$ is same as that of $f'(c) /g'(c)$. – Paramanand Singh Jul 31 '17 at 16:41
• For your example $f, g$ we clearly have $f'(0)/g'(0)=(f(1)-f(-1))/(g(1)-g(-1))=0$ so that Cauchy's MVT is satisfied. – Paramanand Singh Jul 31 '17 at 16:44

I don't think you understand Mean Value Theorem (MVT) correctly (as far as I understood your question).

For $f(x)$ and $g(x)$ MVT gives us the following:

$\exists$ $c_1 \in [a, b]: f^{'}(c_1) = \dfrac{f(b) - f(a)}{b - a}$

$\exists$ $c_2 \in [a, b]: g^{'}(c_2) = \dfrac{g(b) - g(a)}{b - a}$

$\exists$ $c_3 \in [a, b]: \dfrac{f^{'}(c_3)}{g^{'}(c_3)} = \dfrac{f(b) - f(a)}{g(b) - g(a)}$

$c_1, c_2, c_3$ are not related in any way, and MVT gives us no information about the actual values of $c_1, c_2, c_3$, we only know that these points do exist. In addition, $c_1, c_2, c_3$ don't have to be unique.

• For the above functions, where is any such $C_3$? – R004 Jul 31 '17 at 15:00
• What do you mean? $c_3$ is some point $\in [a, b]$, such that this equality holds $\dfrac{f^{'}(c_3)}{g^{'}(c_3)} = \dfrac{f(b) - f(a)}{g(b) - g(a)}$. We don't know where exactly on $[a, b]$ this point is located, we only know that it must exist. – Fedor Jul 31 '17 at 15:05
• I think, $\frac{f'(c_3)}{g'(c_3)}=\frac{\frac{f(b)-f(a)}{b-a}}{\frac{g(b)-g(a)}{b-a}} = \frac{f'(c_1)}{g'(c_2)}$. What does this tell us now? – R004 Jul 31 '17 at 15:10
• Considering my functions, there is no such $c_3$. – R004 Jul 31 '17 at 15:13
• For the functions $f(x) = x^2$ and $g(x) = e^x - 1$, we are getting $\dfrac{f^{'}(c)}{g^{'}(c)} = \dfrac{1^2 - (-1)^2}{e^1 - e^{-1}} = 0$. Although in this case it is easy to see that $c = 0$, (since $f^{'}(0) = 0$), this is not the corollary of MVT, MVT gives us no information about the value of $c$. – Fedor Jul 31 '17 at 15:14