Evaluating limits

$\displaystyle \lim_{n \to \infty} n \tan \left( \dfrac1n\right) = \lim_{n \to \infty} \frac {\tan \left( \dfrac1n\right)}{\dfrac1n} = \dfrac00$

Apply L'Hopital's Rule :

$\displaystyle \lim_{n \to \infty} \frac {\tan \left( \dfrac1n\right)}{\dfrac1n} = \lim_{n \to \infty} \frac {-\dfrac{1}{n^2} \times \sec^2\left(\dfrac1n\right)}{-\dfrac{1}{n^2}} = \lim_{n \to \infty} \sec^2\left(\dfrac1n\right)$

$\displaystyle \lim_{n \to \infty} \sec^2 (0) = \lim_{n \to \infty} \left( \dfrac {1}{\cos 0}\right)\left( \dfrac {1}{\cos 0}\right) = \lim_{n \to \infty} \dfrac 11 \times \dfrac 11 = \lim_{n \to \infty} 1 = 1$

Here when we put $m = \frac{1}{n}$

We get $\displaystyle \lim_{m\to 0} \frac{1}{m} \tan m$

We get a $\frac{0}{0}$ form. Hence using L hospitals rule we get

$\displaystyle \lim_{m\to 0} \frac{sec^2 m}{1}$

Putting m = 0

We get 1. (Answer)