short explanation of l'Hôpital's rule

Dec 2015
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Given limit \(\displaystyle \lim_{x\rightarrow \infty} \frac{x}{\sqrt{1+x^2 }}\), applying l'Hôpital's rule it repeats itself but also turns into an equation.

\(\displaystyle 0<L=\lim_{x\rightarrow \infty} \frac{x'}{(\sqrt{1+x^2 })'}=\lim_{x\rightarrow \infty } \frac{\sqrt{1+x^2 }}{x}=\frac{1}{L} \; \) ; \(\displaystyle \; L^2 =1 \: \Rightarrow L=1\).

Also if the limit involves the factorial then l'Hôpital's rule can make it easier since \(\displaystyle x!=\Gamma (x+1) \approx x^x e^{-x} \sqrt{2\pi x } \; \) , as \(\displaystyle x\rightarrow \infty\).

Example : \(\displaystyle l=\lim_{n\rightarrow \infty} \frac{n^n }{n!} =\lim_{n\rightarrow \infty} \frac{n^n }{\Gamma(n+1) } = \lim_{n\rightarrow \infty} \frac{n^n }{n^n e^{-n} \sqrt{2\pi n }}=\lim_{n\rightarrow \infty} \frac{e^n }{\sqrt{2\pi n }}\).
\(\displaystyle l=\frac{1}{\sqrt{2\pi }}\cdot \lim_{n=t^2 \rightarrow \infty} \frac{(e^{t^2})'}{t'}=\frac{1}{\sqrt{2\pi }}\cdot \lim_{n=t^2 \rightarrow \infty} 2te^{t^2} =\infty\).
 
Dec 2015
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169
Earth
Another one that turns out to be an equation:
$L= \displaystyle\lim_{x \to \infty}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} = \lim_{x \to \infty}\frac{1-e^{-2x}}{1+e^{-2x}} = \frac{1+0}{1+0} = 1 $.
With l'Hôpital's rule: \(\displaystyle 0\leq L=



\lim_{x \to \infty}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=
\lim_{x \to \infty}\frac{(e^{x}-e^{-x})'}{(e^{x}+e^{-x})'}=
\lim_{x \to \infty}\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=\frac{1}{L} \; \Rightarrow L=1.\)
 

romsek

Math Team
Sep 2015
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why does L'Hopital's rule work in general?
 
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Dec 2015
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for \(\displaystyle \lim_{x\rightarrow a } \dfrac{f(x)}{g(x)}=\dfrac{0 }{0}\) then \(\displaystyle f(x)=f(x)-f(a)=(x-a)f'(a).\) same for g(x) .