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.
 A: 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, $$\tan'(\pi/3)$$
A: $$\begin{align}\lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\sqrt 3\right)&=\lim_{n\to\infty}n\left(\frac{\sqrt 3+\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}-\sqrt 3\right)\\&=\lim_{n\to\infty}\frac{4n\tan\frac 1n}{1-\sqrt 3\tan\frac 1n}\\&=\lim_{n\to\infty}\frac{4\tan\frac 1n/(1/n)}{1-\sqrt 3\tan\frac 1n}\\&=\frac{4\cdot 1}{1-0}\end{align}$$
A: Taylor expansion would be a good idea. Around $x=0$ $$\tan(a+x)=\tan (a)+x \left(\tan ^2(a)+1\right)+x^2 \left(\tan ^3(a)+\tan
   (a)\right)+O\left(x^3\right)$$ Replace $x$ by $\frac 1n$ and get $$\tan(a+\frac 1n)-\tan (a)=\left(\tan ^2(a)+1\right)\frac 1n+\cdots$$
A: $$\lim_{n\to \infty}n\left(\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3\right)$$
$$=\lim_{n\to \infty}\frac{\tan\left(\frac{\pi}{3}+\frac{1}{n}\right)-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{\frac{\tan\frac{\pi}{3}+\tan \frac{1}{n}}{1-\tan \frac{\pi}{3}\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{\frac{\sqrt 3+\tan \frac{1}{n}}{1-\sqrt 3\tan \frac{1}{n}}-\sqrt 3}{\frac{1}{n}}$$
$$=\lim_{n\to \infty}\frac{4\tan\frac{1}{n}}{\frac{1}{n}(1-\sqrt 3\tan\frac{1}{n})}$$
$$=4\lim_{n\to \infty}\frac{\tan\frac{1}{n}}{\frac{1}{n}}\times \lim_{n\to \infty}\frac{1}{1-\sqrt 3\tan \frac{1}{n}}$$
$$=4(1)\left(\frac{1}{1-0}\right)=\color{red}{4}$$
A: \begin{align}
\lim_{n\to\infty}n\left(\tan\left(\frac{\pi}{3}+\frac 1n\right)-\tan\frac{\pi}{3} \right)&=\lim_{n\to\infty}n\left(\tan(\frac{\pi}{3}+\frac 1n-\frac{\pi}{3})\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big) \right)\\
&=\lim_{n\to\infty}\frac{n\sin\frac 1n}{\cos \frac 1n}\times \lim_{n\to\infty}\Big(1+\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\
&=\frac{\lim_{n\to\infty}n\sin\frac 1n}{\lim_{n\to\infty}\cos \frac 1n}\times \Big(1+\lim_{n\to\infty}\tan(\frac{\pi}{3}+\frac 1n)\tan\frac{\pi}{3}\Big)\\
&=\frac{1}{1}\times(1+\sqrt 3\sqrt 3)\\
&=4
\end{align}
A: It's very tempting to use L'Hopital's rule, indeed. An alternative proof may be like this: we construct approximations to the function $f (n) = \tan (\pi/3 + 1/n)$. For example, Taylor Series Approximation gives $$\tan \left (\frac {\pi} {3} + \frac {1} {n} \right) = \tan \frac {\pi} {3} + \frac {4} {n} + o \left (\frac {1} {n}\right).$$ Putting this back in the limit we have $$\lim_{n \to \infty} n \left(\tan \left(\frac {\pi} {3} + \frac {1} {n} \right) - \sqrt{3} \right) = \lim_{n \to \infty} (4 + o (1)) = 4.$$
