Limit of $n \tan(\frac{x}{n})$

Question

Let $f_n :[-1/2, 1/2] \to \mathbb{R}$, be a seq. of functions, where $f_n(x) = n \tan(\frac{x}{n})$. Im asked to show that this sequence converges pointwise and determine the limit function. I have no idea where to even start.

It seems that $\forall x: f_n(x) \to x$

Any hints?

I know that $\cos(x/n) \to 1$ and that $\sin(x/n)$ behaves like $x/n$ as $n$ increases. So it makes "sense" that the limit function is $x$, but I'm failing to show it rigorously.

Answer 1

Notice that

$$n\tan\left(\frac{x}{n}\right)=n\frac{\sin\left(\frac{x}{n}\right)}{\cos\left(\frac{x}{n}\right)}=x\cdot\underbrace{\frac{\sin\left(\frac{x}{n}\right)}{x/n}}_{\to1}\cdot\underbrace{\frac{1}{\cos\left(\frac{x}{n}\right)}}_{\to1}\to x$$

as $n\to\infty$.

Answer 2

From the hint on the comment section: $$n \tan\left(\frac{x}{n}\right) = x \frac{\tan(x/n)}{x/n}$$

Since $\frac{\tan(x)}{x} \to 1$ as $x \to 0$, we get that $f_n(x)\to x$

Answer 3

I like to use L'hopital when I can: $$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}n\tan\left(\dfrac xn\right)=\lim_{n\to\infty}\dfrac {\tan\left(\dfrac xn\right)}{\dfrac 1n}\overset {\text {L'hopital}}=\lim_{n\to\infty}\dfrac {\dfrac {-x}{n^2}\sec^2\left(\dfrac xn\right)}{-\dfrac 1{n^2}}=\lim_{n\to\infty}x\sec^2\left(\dfrac xn\right)=x\cdot\sec^2(0)=x\cdot 1=x$$.