Compute lim without using L'Hospital's rule.
By using L'Hospital's rule and
\tan'( \Diamond )=( \Diamond )'(1+\tan^{2}( \Diamond ))
I mean by \Diamond a function
so I got
\begin{align} \lim_{n\to +\infty}n\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right) &=\lim_{n\to +\infty}\dfrac{\left(\tan\left(\dfrac{\pi}{3}+\dfrac{1}{n} \right)-\sqrt{3}\right)}{\dfrac{1}{n}}\\ &=1+\tan^{2}\left(\dfrac{\pi}{3}\right)=1+\sqrt{3}^{2}=1+3=4 \end{align}
I'm interested in more ways of computing limit for this sequence.
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