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# Try to solve it 1

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

Note by A K
1 year, 1 month 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 $$u \rightarrow 0.$$ · 1 year, 1 month ago

Nice! Approaches without using L'hopital's rule are always really good. · 1 year, 1 month ago

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. :) · 1 year, 1 month ago

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. · 1 year, 1 month ago

-4π/3 · 4 months ago

$$\frac{-4\pi}{3}$$ · 1 year, 1 month ago