Skip to content

12.23 More theorems of derivates

Generalized Mean Value Theorem (Cauchy Mean Value Theorem)

Let continuous \(f,g:[a,b]\rightarrow \R\) be differentiable on \((a,b)\). Then \(\exists c\in (a,b)\) such that \(f^{\prime}(c)\left(g\left(b\right)-g\left(a\right)\right)=g^{\prime}\left(c\right)\left(f\left(b\right)-f\left(a\right)\right)\)​

In particular \(g'(c)\neq 0 \wedge (g(b)-g(a)\neq 0)\), then we have \(\frac{f^{\prime}(c)}{g^{\prime}\left(c\right)}=\frac{f\left(b\right)-f\left(a\right)}{g\left(b\right)-g\left(a\right)}\)

Remark

  • Lagrange Mean Value Theorem is a particular case of this one for \(g(x)=x\)
  • If we apply MVT to \(f\) and \(g\), we obtain \(\begin{cases}\exists c_1\in(a,b):f^{\prime}(c_1)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ \exists c_2\in(a,b):g^{\prime}(c_2)=\frac{g\left(b\right)-g\left(a\right)}{b-a}\end{cases}\Rightarrow\frac{f^{\prime}(c_1)}{g^{\prime}\left(c_2\right)}=\frac{f\left(b\right)-f\left(a\right)}{g\left(b\right)-g\left(a\right)}\)​

Then we need to get the same \(c\) and consider the situation that the denominator is zero

Proof

Consider \(h(x)=(g(b)-g(a))f(x)-(f(b)-f(a))g(x)\)

Then \(h(a)=g(b)f(a)-f(b)g(a)\) and \(h(b)=-g(a)f(b)+f(a)g(b)=h(a)\)

Then by Rolle's Theorem, \(\exists c\in (a,b)\) such that \(h'(c)=0\)

(Note that \(h\) is differentiable on \((a,b)\) and continuous on \([a,b]\) and \(h(a)=h(b)\))

But \(h^{\prime}(x)=f^{\prime}(x)\left(g\left(b\right)-g\left(a\right)\right)-g^{\prime}\left(x\right)\left(f\left(b\right)-f\left(a\right)\right)\), since \(h'(c)=0\)

We get \(f^{\prime}(c)\left(g\left(b\right)-g\left(a\right)\right)=g^{\prime}\left(c\right)\left(f\left(b\right)-f\left(a\right)\right)\)

L'Hopital's Rule

  • First version (\(0/0\) case)

Let \(f,g:(a-\delta_0,a+\delta_0)-\{a\}\rightarrow \R\) be such that \(f,g\) are differentiable on \((a-\delta_0,a+\delta_0)-\{a\}\) and \(g'(x)\neq 0,\forall x\in (a-\delta_0,a+\delta_0)-\{a\}\)​

If \(\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=0\) and \(\lim_{x\to a}\frac{f^{\prime}\left(x\right)}{g^{\prime}\left(x\right)}=l\), then \(\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=l\)

Summarizing: How to figure out the hypothesis?

Since we take the limit of \(\frac{f(x)}{g(x)}\)when \(x\to a\)\, then we don't care \(f(a),g(a)\), thus we need \((a-\delta_0,a+\delta_0)-\{a\}\)

Also, the need \(\lim_{x\to a}\frac{f^{\prime}\left(x\right)}{g^{\prime}\left(x\right)}=l\), then the differentiable should be defined and \(g'(x)\neq 0\)​

Proof

We want to use GMVT that is we need a closed interval instead of \((a-\delta_0,a+\delta_0)-\{a\}\)

Let's define \(f(a)=g(a)=0\). Now \(f,g\) are continuous on \((a-\delta_0,a+\delta_0)\) since \(\lim_{x\to a}f\left(x\right)=f\left(a\right)=0,\lim_{x\to a}g\left(x\right)=g\left(a\right)=0\)

Moreover

  • If \(x\in (a-\delta_0,a)\), then \(f,g:[x,a]\to \R\) are continuous and differentiable on \((x,a)\)

    We use GMVT here, \(\exists\alpha\in(a-\delta_0,a)\) such that \(\frac{f^{\prime}(\alpha)}{g^{\prime}\left(\alpha\right)}=\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}=\frac{f\left(x\right)}{g\left(x\right)}\) where \(g(x)\neq 0\)

    Why \(g(x)\neq 0\), because if \(g(x)=0=g(a)\), then by rolle's theorem \(\exists c\in (x,a)\) such that \(g'(c)=0\) contradicts to hypothesis.

    Again GMVT, \(\exists\alpha_{x}\in(x,a)\) where \(\alpha_x=\alpha(x)\) depends on \(x\) such that \(\frac{f^{\prime}(\alpha_{x})}{g^{\prime}\left(\alpha_{x}\right)}=\frac{f\left(x\right)}{g\left(x\right)}\)

    Clearly, if \(x\to a^-\), then \(\alpha_x\to a\) since \(x<\alpha_x<a\) * If \(x\in (a,a+\delta_0)\), then \(f,g:[a,x]\to\mathbb{R}\) are continuous and differentiable on \((a,x)\)​

    We use GMVT here, \(\exists\beta_{x}\in(a,x)\) such that \(\frac{f^{\prime}(\beta_{x})}{g^{\prime}\left(\beta_{x}\right)}=\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}=\frac{f\left(x\right)}{g\left(x\right)}\) where \(\beta_x\to a\) when \(x\to a\) since \(a<\beta_x<x\)​

Now, \(\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=\begin{cases}\lim_{x\to a^{-}}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x\to a^{-}}\frac{f^{\prime}\left(\alpha_{x}\right)}{g^{\prime}\left(\alpha_{x}\right)}=?\\ \lim_{x\to a^{+}}\frac{f\left(x\right)}{g\left(x\right)}=\lim_{x\to a^{+}}\frac{f^{\prime}\left(\beta_{x}\right)}{g^{\prime}\left(\beta_{x}\right)}=?\end{cases}\) since theorem of composition of limit

Since \(\lim_{x\to a^{-}}\alpha_{x}=a\) and \(\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}\), then \(\lim_{x\to a^{-}}\frac{f^{\prime}\left(\alpha_{x}\right)}{g^{\prime}\left(\alpha_{x}\right)}=\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}\)​

We have \(?=\lim_{x\to a}\frac{f^{\prime}\left(x\right)}{g^{\prime}\left(x\right)}=l\) by hypothesis * Second version (\(+\infty/+\infty\) case)

Let \(f,g:(a-\delta_0,a+\delta_0)-\{a\}\rightarrow \R\) be such that \(f,g\) are differentiable on \((a-\delta_0,a+\delta_0)-\{a\}\) and \(g'(x)\neq 0,\forall x\in (a-\delta_0,a+\delta_0)-\{a\}\)

If \(\lim_{x\to a}f\left(x\right)=\lim_{x\to a}g\left(x\right)=\infty\) and \(\lim_{x\to a}\frac{f^{\prime}\left(x\right)}{g^{\prime}\left(x\right)}=l\), then \(\lim_{x\to a}\frac{f\left(x\right)}{g\left(x\right)}=l\)

Proof

Example

  1. \(\lim_{x\to0}\frac{\sin x}{x}\)

    • Usually this is unnecessary to use L'Hopital's Rule, we can just use squeeze theorem that is \(\frac{1}{\cos x}\) and \(\frac{1}{x}\)
    • Or \(\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\sin\left(x\right)-\sin\left(0\right)}{x-0}\overset{def}{=}\sin^{\prime}\left(0\right)=\cos\left(0\right)=1\)
    • Or \(\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{\cos x}{1}=\frac{\cos\left(0\right)}{1}=1\)

  2. \(\lim_{x\to0}\frac{\left(e^{x}-1\right)x}{\sin^2\left(x\right)}\)

    • Let \(f(x)=\left(e^{x}-1\right)x\) and \(g(x)=\sin ^2(x)\)

    Then \(g'(x)=2\sin x\sin'(x)=2\sin x\cos x\), \(g'(x)=0\) only when \(x=0=a\), thus we can use L'Hopital's Rule

    Then \(\lim_{x\to0}\frac{\left(e^{x}-1\right)x}{\sin^2\left(x\right)}=\lim_{x\to0}\frac{xe^{x}+e^{x}-1}{2\sin x\cos x}=\lim_{x\to0}\frac{xe^{x}+2e^{x}}{2\cos^2\left(x\right)-2\sin^2\left(x\right)}\)

    Since \((2\sin x\cos x)^{\prime}=2\cos^2\left(x\right)-2\sin^2\left(x\right)\) since this is not \(0\) in the interval

    Thus use L'Hopital's Rule again \(\lim_{x\to0}\frac{\left(e^{x}-1\right)x}{\sin^2\left(x\right)}=\frac22=1\) * However, we can simplify the expression initially

    \(\lim_{x\to0}\frac{\left(e^{x}-1\right)x}{\sin^2\left(x\right)}=\lim_{x\to0}\frac{e^{x}-1}{\sin x}\cdot\lim_{x\to0}\frac{x}{\sin\left(x\right)}=\lim_{x\to0}\frac{e^{x}-1}{\sin x}\), then use L'Hopital's Rule just once

    Thus don't use L'Hopital's Rule too quickly

‍