<s>[INST] Compute $\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$ without using L' Hôpital 
Compute $$\lim_{n\to +\infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n} \right)-\sqrt{3}\right)$$ 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.
 [/INST] The expression is a derivative. Since $\sqrt{3} = \tan\left(\frac{\pi}{3}\right)$, the limit is
$$\lim_{n \to \infty} n \left( \tan\left(\frac{\pi}{3} + \frac{1}{n} \right) - \tan\left(\frac{\pi}{3}\right)\right)$$
That is by the substitution $h = \frac{1}{n}$, $$\lim_{h \to 0} \frac{1}{h} \left(\tan\left(\frac{\pi}{3} + h \right) - \tan \left(\frac{\pi}{3} \right)\right)$$
That is, $$\tan'(\pi/3)$$
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