calculus - Compute $lim_{nto +infty}nleft(tanleft(frac{pi}{3}+frac{1}{n} right)-sqrt{3}right)$ 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.