limits - By letting $m=frac{1}{n}$ find $lim_{nrightarrowinfty} n tan left(frac{1}{n}right)$

By letting $m=\dfrac{1}{n}$ find

$$\lim\limits_{n\rightarrow\infty} n \tan \left(\dfrac{1}{n}\right)$$

I've played around with the algebra, but can't see how m fits in, apart from abbreviation. Is my working correct?

$\dfrac{\tan{\frac{1}{n}}}{\frac{1}{n}}=\frac{0}{0}$

L'Hopital's rule:

$\dfrac{\sec^2m}{(-1/n^2 )}=\dfrac{-n^2}{\cos^2\frac{1}{n}}$

Divide by $-n^2$ to get $\dfrac{1}{\cos^2\frac{1}{n}}$ which is 1 as n tends to infinity because $\cos0=1$. I don't really get limits yet, but is this right?

Answer

Here are the steps
$$\lim\limits_{n\to\infty} n\tan\left(\frac{1}{n}\right)$$
Let $m=\frac{1}{n}$. Since $n\to\infty$, then $m=\frac{1}{\infty}=0$. So now we have
$$\lim\limits_{m\to 0} \frac{\tan\left(m\right)}{m}= \lim\limits_{m\to 0} \frac{\frac{d}{dm}\tan\left(m\right)}{\frac{d}{dm}m} = \lim\limits_{m\to 0} \sec^2\left(m\right) = \lim\limits_{m\to 0} \frac{1}{\cos^2\left(m\right)}=1$$
Looks to me like you do get limits. Keep it up.