By letting $m=\frac{1}{n}$ find $\lim_{n\rightarrow\infty} n \tan \left(\frac{1}{n}\right)$ 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? 
 A: $$\lim_{n\rightarrow\infty} n \tan (\dfrac{1}{n})\tag{1}$$
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:
$$\lim_{m\rightarrow 0} \frac{1}{m} \tan (m)=1\tag{3}$$
A: Your work is perfectly fine. You "do get limits" in the end since you've reduced the question to calculating the limit $\lim_{n\rightarrow \infty} \frac{1}{\cos^2(1/n)}=1$. 
The suggestion is that $m=1/n$ is a trick to make things look cleaner. It says that the original question is equivalent to evaluating $\lim_{m\rightarrow\ 0 } \frac{\tan(m)}{m}$, and you can do this with either l'Hopital's rule or by writing $\tan(m)=\sin(m)/\cos(m)$ and noticing $\cos(m)\rightarrow 1$, and $\sin(m)/m\rightarrow 1$ so the limit is 1 (perhaps the latter you know already). The point is this "trick" makes it immediately obvious that a problem involving $1/n$ with $n\rightarrow\infty$ can be rewritten as $m\rightarrow 0$ with $m=1/n$. 
A: Without differentials, l'Hospital and stuff:
$$\lim_{n\to\infty}n\tan\frac1n=\lim_{n\to\infty}\frac{\tan\frac1n}{\frac1n}\stackrel{m=\frac1n}=\lim_{m\to 0}\frac{\sin m}m\cdot\frac1{\cos m}=1\cdot 1=1$$
A: As $n \to \infty,$ note $x=1/n \to 0.$ So using LHopital on $(\tan x )/x$ gives just $\sec^2 x/1$ which immediately approaches $1.$
A: 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.
