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L'Hospital's rule and Taylor's theorem

  1. If \(f\) be defined on \([a,b]\) and for \(x\in[a,b]\) the limit \[f'(x)=\underset{t\to x}{\lim}\frac{f(t)-f(x)}{t-x}\] exists, we say that \(f\) is differentiable at \(x\). If \(f'\) is defined at every point of a set \(E\subset[a,b]\), we say that \(f\) is differentiable on \(E\).

  2. The mean value theorem If \(f\) and \(g\) are continuous real functions on \([a,b]\) which are differentiable in \((a,b)\), then there is a point \(x\in(a,b)\) at which \[\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(x)}{g'(x)}\] Put \(h(t)=[f(b)-f(a)]g(t)-[g(b)-g(a)]f(t)\quad(a\le t\le b)\) then \(h\) is continuous on \([a,b]\) and differentiable in \((a,b)\), and \(h(a)=[f(b)-f(a)]g(a)-[g(b)-g(a)]f(a)=f(b)g(a)-g(b)f(a)=h(b)\) To prove this theorem, we have to show that \(h'(x)=0\) for some \(x\in(a,b)\). If \(h\) is constant, this holds for every \(x\in(a,b)\). If \(h(t)>h(a)\) for some \(t\in(a,b)\), let \(x\) be a point on \([a,b]\) at which \(h\) attains its maximum, then \(h'(x)=0\). If \(h(t)<h(a)\) for some \(t\in(a,b)\), the same argument applies if we choose for \(x\) a point on \([a,b]\) where \(h\) attains its minimum and \(h'(x)=0\).

  3. L’Hospital’s rule: Suppose \(f\) and \(g\) are real and differentiable in \((a, b)\) and \(g'(x)\ne 0\) for all \(x\in (a,b)\) where \(-\infty\le a<b\le+\infty\). Suppose \(\displaystyle\frac{f'(x)}{g'(x)}\to A\) as \(x\to a\). Suppose \(-\infty\le A<+\infty\), choose a real number \(q\) such that \(A<q\) and choose \(r\) such that \(A<r<q\), there is point \(c\in(a,b)\) such that \(a<x<c\) implies \[\frac{f'(x)}{g'(x)}<r\]

  1. Suppose \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a\). If \(a<x<y<c\), then there is point \(t\in (x,y)\) such that \[\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(t)}{g'(t)}<r\] Suppose \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a\), then \[\frac{f(y)}{g(y)}\le r<q\quad (a<y<c)\] Then \[\frac{f(y)}{g(y)}\to A\] or \[\frac{f(x)}{g(x)}\to A\] as \(x\to a\)
  2. Suppose \(g(x)\to +\infty\) as \(x\to a\). If \(a<x<y<c\), then there is point \(c_1\in (a,y)\) such that \(g(x)>g(y)\) and \(g(x)>0\) if \(a<x<c_1\) then \[\frac{f(x)-f(y)}{g(x)-g(y)}\frac{g(x)-g(y)}{g(x)}=\frac{f(x)-f(y)}{g(x)}<r\frac{g(x)-g(y)}{g(x)}=r-r\frac{g(y)}{g(x)}\] and then \[\frac{f(x)}{g(x)}<r-r\frac{g(y)}{g(x)}+\frac{f(y)}{g(x)}, \quad (a<x<c_1)\] Because \(g(x)\to +\infty\) as \(x\to a\), then there is point \(c_2\in (a,c_1)\) \[\frac{f(x)}{g(x)}<r<q, \quad (a<x<c_2)\] and \[\frac{f(x)}{g(x)}\to A\] as \(x\to a\).
  3. In the same manner, if \(-\infty<A\le+\infty\) and \(p\) is chosen so that \(p<A\), we can find a point \(c_3\) such that \[p<\frac{f(x)}{g(x)},\quad(a<x<c_3)\] and \[\frac{f(x)}{g(x)}\to A\] as \(x\to a\).
  1. Taylor’s theorem: Soppose \(f\) is a real function on \([a,b]\), \(n\) is a positive integer, \(f^{n-1}\) is continuous on \([a,b]\). Let \(\alpha\), \(\beta\) be distinct points of \([a,b]\) then there exist a point \(x\) between \(\alpha\) and \(\beta\) such that \[f(\beta)=\underset{P(\beta)}{\underbrace{\frac{f^{(0)}(\alpha)}{0!}(\beta-\alpha)^0+\frac{f^{(1)}(\alpha)}{1!}(\beta-\alpha)^1+\cdots+\frac{f^{(n-1)}(\alpha)}{(n-1)!}(\beta-\alpha)^{n-1}}}+\frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n\] Let \(M\) be the number defined by \[f(\beta)=P(\beta)+M(\beta-\alpha)^n\] and let \[g(t)=f(t)-P(t)-M(t-\alpha)^n\quad(a\le t\le b)\] and \[g^{(n)}(t)=f^{(n)}(t)-n!M\quad(a<t<b)\] Since \(g(\alpha)=f(\alpha)-P(\alpha)=0\) and our choice of \(M\) shows that \(g(\beta)=f(\beta)-P(\beta)-M(\beta-\alpha)^n=0\), so that \(g'(x_1)=0\) for some \(x_1\) between \(\alpha\) and \(\beta\). Since \(g'(\alpha)=f'(\alpha)-P'(\alpha)-nM(\alpha-\alpha)^{n-1}=0\), we conclude that \(g''(x_2)=0\) for some \(x_2\) between \(\alpha\) and \(x_1\). After \(n\) steps we arrive at the conclusion that \(g^{(n)}(x_n)=0\) for some \(x_n\) between \(\alpha\) and \(x_{n-1}\) that is, between \(\alpha\) and \(\beta\). Then \(f^{(n)}(x)=n!M\) when \(x=x_n\) and \[f(\beta)=\underset{P(\beta)}{\underbrace{\frac{f^{(0)}(\alpha)}{0!}(\beta-\alpha)^0+\frac{f^{(1)}(\alpha)}{1!}(\beta-\alpha)^1+\cdots+\frac{f^{(n-1)}(\alpha)}{(n-1)!}(\beta-\alpha)^{n-1}}}+\frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n\]