Try to solve it 1

limx133tan(πx)3x1= ? \large \lim_{x\to\frac13} \dfrac{\sqrt3 - \tan(\pi x)}{3x-1} = \ ?

#Calculus #Limits

Note by Abdulrahman El Shafei
5 years, 8 months ago

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Comments

We could just use L'Hopital's rule, but I'm assuming that you are looking for an approach that does not use this rule.

Let u=x13.u = x - \dfrac{1}{3}. Then u0u \rightarrow 0 as x13x \rightarrow \dfrac{1}{3} and x=u+13,x = u + \dfrac{1}{3}, in which case

3tan(πx)=3tan(πu+π3)=3tan(πu)+tan(π3)1tan(πu)tan(π3)=\sqrt{3} - \tan(\pi x) = \sqrt{3} - \tan\left(\pi u + \dfrac{\pi}{3}\right) = \sqrt{3} - \dfrac{\tan(\pi u) + \tan(\frac{\pi}{3})}{1 - \tan(\pi u)\tan(\frac{\pi}{3})} =

3tan(πu)+313tan(πu)=33tan(πu)tan(πu)313tan(πu)=4tan(πu)13tan(πu).\sqrt{3} - \dfrac{\tan(\pi u) + \sqrt{3}}{1 - \sqrt{3}\tan(\pi u)} = \dfrac{\sqrt{3} - 3\tan(\pi u) - \tan(\pi u) - \sqrt{3}}{1 - \sqrt{3}\tan(\pi u)} = \dfrac{-4\tan(\pi u)}{1 - \sqrt{3}\tan(\pi u)}.

Thus the original limit can be written as

limu04tan(πu)3u(13tan(πu))=\lim_{u \rightarrow 0} \dfrac{-4\tan(\pi u)}{3u*(1 - \sqrt{3}\tan(\pi u))} =

limu04πtan(πu)3(πu)limu0113tan(πu)=4π3limw0tan(w)w1=4π3,\lim_{u \rightarrow 0} \dfrac{-4\pi \tan(\pi u)}{3*(\pi u)} * \lim_{u \rightarrow 0} \dfrac{1}{1 - \sqrt{3}\tan(\pi u)} = -\dfrac{4\pi}{3} * \lim_{w \rightarrow 0} \dfrac{\tan(w)}{w} * 1 = \boxed{-\dfrac{4 \pi}{3}},

where w=πu0w = \pi u \rightarrow 0 as u0.u \rightarrow 0. Note also that (13tan(πu))1(1 - \sqrt{3}\tan(\pi u)) \rightarrow 1 as u0.u \rightarrow 0.

Brian Charlesworth - 5 years, 8 months ago

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Nice! Approaches without using L'hopital's rule are always really good.

Kartik Sharma - 5 years, 8 months ago

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Thanks! L'Hopital's is really useful if you want a quick result, but I like the challenge of trying to find an alternative approach. :)

Brian Charlesworth - 5 years, 8 months ago

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@Brian Charlesworth Another way is to put πx=y \pi x = y and by first principles we can say the limit equals negative of the derivative of tany \tan y at y=π3 y = \dfrac{ \pi}{3} . This is similar to L'Hospital but not completely.

Sudeep Salgia - 5 years, 8 months ago

4π3\frac{-4\pi}{3}

Aditya Kumar - 5 years, 8 months ago

-4π/3

Jahid Rafi - 4 years, 11 months ago
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