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?
Best Answer
$$\lim_{n\rightarrow\infty} n \tan (\dfrac{1}{n})\tag{1}$$
Setting $m=1/n$, (1) becomes
$$\lim_{m\rightarrow 0} \frac{1}{m} \tan (m)\tag{2}$$
When $m\to 0$, we have $\tan(m)\to m+m^3/3+...$. Thus we obtain:
$$\lim_{m\rightarrow 0} \frac{1}{m} \tan (m)=1\tag{3}$$