Generalisations of l'Hospital's Rule
Further scenarios in which limits can be calculated by l'Hospital's rule.
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Study Generalisations of l’Hospital’s Rule
Generalisations of l’Hospital’s Theorem
Expressions of type \(\frac\infty\infty\), Limit as \(x\to x_0\)
Let \(f,g:I\backslash\{x_0\}\to\mathbb{R}\) be differentiable functions with \(\lim_{x\to x_0}f(x)=\infty\), \(\lim_{x\to x_0}g(x)=\infty\). 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 with \[\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{f'(x)}{g'(x)}.\]Expressions of type \(\frac00\), Limit as \(x\to\infty\)
Let \(f,g:[t_0,\infty)\to\mathbb{R}\) be differentiable functions with \(\lim_{x\to \infty}f(x)=0\), \(\lim_{x\to \infty}g(x)=0\). Assume that there exists some \(x_1\in\mathbb{R}\) such that \(g'(x)\neq0\) for all \(x>x_1\). Then, if \(\lim_{x\to\infty}\frac{f'(x)}{g'(x)}\) exists, then also \(\lim_{x\to\infty}\frac{f(x)}{g(x)}\) exists with \[\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}.\]Expressions of type \(\frac\infty\infty\), Limit as \(x\to\infty\)
Let \(f,g:[t_0,\infty)\) be differentiable functions with \(\lim_{x\to \infty}f(x)=\infty\), \(\lim_{x\to \infty}g(x)=\infty\). Then, if \(\lim_{x\to\infty}\frac{f'(x)}{g'(x)}\) exists, then also \(\lim_{x\to\infty}\frac{f(x)}{g(x)}\) exists with \[\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty}\frac{f'(x)}{g'(x)}.\]
Proof: The first result follows by an application of l’Hospital’s Theorem to \[\lim_{x\to x_0}\frac{f(x)}{g(x)}=\lim_{x\to x_0}\frac{\frac1{g(x)}}{\frac1{f(x)}}=\lim_{x\to x_0}\frac{\frac{g'(x)}{(g(x))^2}}{\frac{f'(x)}{(f(x))^2}}=\frac1{\lim_{x\to x_0}\frac{f'(x)}{g'(x)}}\left(\lim_{x\to x_0}\frac{f(x)}{g(x)}\right)^2.\] The second and third statement follow by a substitution \(y=\frac1x\) and the consideration of \[\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{y\to0}\frac{f\left(\frac1y\right)}{g\left(\frac1y\right)} =\lim_{y\to0}\frac{-\frac1{y^2}f'\left(\frac1y\right)}{-\frac1{y^2}g'\left(\frac1y\right)} =\lim_{y\to0}\frac{f'\left(\frac1y\right)}{g'\left(\frac1y\right)} =\lim_{x\to\infty}\frac{f'(x)}{g'(x)}.\] \(\Box\)
Note that we can also treat expressions of type “\(\infty-\infty\)” by l’Hospitals’s Theorem. Namely, for \(f,g\) with \(\lim_{x\to x_0}f(x)=0\), \(\lim_{x\to x_0}g(x)=0\), we get that \[\lim_{x\to x_0}(\frac{1}{f(x)}-\frac{1}{g(x)})=\lim_{x\to x_0}\frac{g(x)-f(x)}{f(x)\cdot g(x)}=\lim_{x\to x_0}\frac{g'(x)-f'(x)}{f'(x)\cdot g(x)+f(x)\cdot g'(x)}.\] Also, expressions of type “\(0\cdot\infty\)” can be treated by a special trick. Namely, for \(f,g\) with \(\lim_{x\to x_0}f(x)=0\), \(\lim_{x\to x_0}g(x)=\infty\), we get that \[\lim_{x\to x_0}f(x)\cdot g(x)=\lim_{x\to x_0}\frac{f(x)}{\frac1{g(x)}}=\lim_{x\to x_0}\frac{f'(x)}{-\frac{g'(x)}{(g(x))^2}}=-\lim_{x\to x_0}\frac{f'(x) (g(x))^2}{g'(x)}.\] You do not have to keep the above two formulas in mind. These can always be derived in concrete examples.
Let \(n\in\mathbb{N}\) and consider \[\lim_{x\to\frac\pi2,x<\frac\pi2}\left(x-\frac\pi2\right)^n\cdot\tan(x).\] This is an expression of type “\(0\cdot\infty\)” and we can make use of \[\begin{aligned} \lim_{x\to\frac\pi2,x<\frac\pi2}\left(x-\frac\pi2\right)^n\cdot\tan(x)=&\lim_{x\to\frac\pi2,x<\frac\pi2}\frac{\left(x-\frac\pi2\right)^n}{\frac1{\tan(x)}}\\=& \lim_{x\to\frac\pi2,x<\frac\pi2}\frac{n\cdot\left(x-\frac\pi2\right)^{n-1}}{-\frac1{\tan^2(x)}\frac1{\cos^2(x)}}\\=& \lim_{x\to\frac\pi2,x<\frac\pi2}\frac{n\cdot\left(x-\frac\pi2\right)^{n-1}}{-\frac1{\sin^2(x)}}\\ =&-\lim_{x\to\frac\pi2,x<\frac\pi2}n\cdot\left(x-\frac\pi2\right)^{n-1}\sin^2(x)\\ =&\begin{cases}-1&\text{ if }n=1,\\0&\text{ else.}\end{cases} \end{aligned}\]
In this example, we first need to bring both terms into a common fraction \[\begin{aligned}\lim_{x\to0}\left(\frac1{\sin(x)}-\frac1x\right)=& \lim_{x\to0}\frac{x-\sin(x)}{x\sin(x)}\\=& \lim_{x\to0}\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\\=& \lim_{x\to0}\frac{\sin(x)}{\cos(x)-x\sin(x)+\cos(x)}=0. \end{aligned}\]
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