Theorem of l'Hospital
An important tool for calculating limits.
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Study Theorem of l’Hospital
Now we consider a generalisation of the mean value theorem. This gives us a completely new tool for the determination of limits.
Theorem 1 (Generalised mean value theorem). Let \(f,g:[a,b]\to\mathbb{R}\) be differentiable. Assume that \(g'\) has no zero in \((a,b)\). Then there exists some \(\hat{x}\in(a,b)\) such that \[\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(\hat{x})}{g'(\hat{x})}.\]
Proof: By introducing the function \[F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}\cdot(g(x)-g(a)),\] we get \(F(a)=F(b)=0\). Now using the Theorem of Rolle, the result follows immediately.\(\Box\)
Theorem 2 (Theorem of l’Hospital). Let \(I\) be an interval and let \(f,g:I\to\mathbb{R}\) be differentiable. Let \(x_0\in I\) and assume that \(f(x_0)=g(x_0)=0\) and there exists some neighbourhood \(U\subset I\) of \(x_0\) such that \(g'(x)\neq0\) for all \(x\in U\backslash\{x_0\}\). Then, if \(\lim_{x\to x_0}\frac{f'(x)}{g'(x)}\) exists, then also \(\lim_{x\to x_0}\frac{f(x)}{g(x)}\) exists and \[\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}.\]
Proof: Let \((x_n)_{n\in\mathbb{N}}\) be a sequence with
\(\lim_{n\to\infty}x_n=x_0\) and \(x_n\neq x_0\) for all \(n\in\mathbb{N}\). Then, by the generalised
mean value theorem, there exists a sequence \((\hat{x}_n)_{n\in\mathbb{N}}\) with \(\hat{x}_n\) between \(x_0\) and \(x_n\) such that \[\frac{f(x_n)}{g(x_n)}=\frac{f(x_n)-f(x_0)}{g(x_n)-g(x_0)}=\frac{f'(\hat{x}_n)}{g'(\hat{x}_n)}.\]
In particular, since \((\hat{x}_n)_{n\in\mathbb{N}}\) converges to
\(x_0\), we have \[\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to
x_0}\frac{f'(x)}{g'(x)}.\] \(\Box\)
Example 3.
Let \(a\in\mathbb{R}\). \[\lim_{x\to0,ax>0}\frac{\log(1+ax)}x=\lim_{x\to0,ax>0}\frac{a}{1+ax}=a.\]
\[\lim_{x\to0,ax>0}(1+ax)^{\frac{1}x}=\lim_{x\to0,ax>0}\exp\left(\frac{\log(1+ax)}x\right)=\exp\left(\lim_{x\to0,ax>0}\frac{\log(1+ax)}x\right)=\exp(a).\] In particular, for \(x=\frac1n\), we have \[\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^{n}=\exp(a).\]
\[\lim_{x\to0}\frac{1-\cos(x)}{x^2}=\lim_{x\to0}\frac{\sin(x)}{2x}=\lim_{x\to0}\frac{\cos(x)}{2}=\frac12.\]
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