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\[ \large \lim_{x\to\frac13} \dfrac{\sqrt3 - \tan(\pi x)}{3x-1} = \ ? \]

Note by A K 2 years, 9 months ago

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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 \(u = x - \dfrac{1}{3}.\) Then \(u \rightarrow 0\) as \(x \rightarrow \dfrac{1}{3}\) and \(x = u + \dfrac{1}{3},\) in which case

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

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

Thus the original limit can be written as

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

\(\lim_{u \rightarrow 0} \dfrac{-4\pi \tan(\pi u)}{3*(\pi u)} * \lim_{u \rightarrow 0} \dfrac{1}{1 - \sqrt{3}\tan(\pi u)} = -\dfrac{4\pi}{3} * \lim_{w \rightarrow 0} \dfrac{\tan(w)}{w} * 1 = \boxed{-\dfrac{4 \pi}{3}},\)

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

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

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

-4π/3

\(\frac{-4\pi}{3}\)

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## Comments

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TopNewestWe 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 \(u = x - \dfrac{1}{3}.\) Then \(u \rightarrow 0\) as \(x \rightarrow \dfrac{1}{3}\) and \(x = u + \dfrac{1}{3},\) in which case

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

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

Thus the original limit can be written as

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

\(\lim_{u \rightarrow 0} \dfrac{-4\pi \tan(\pi u)}{3*(\pi u)} * \lim_{u \rightarrow 0} \dfrac{1}{1 - \sqrt{3}\tan(\pi u)} = -\dfrac{4\pi}{3} * \lim_{w \rightarrow 0} \dfrac{\tan(w)}{w} * 1 = \boxed{-\dfrac{4 \pi}{3}},\)

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

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

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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. :)

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-4π/3

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\(\frac{-4\pi}{3}\)

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