Why is:
$$ \lim_{n\to\infty}n \arctan\left(\frac{x}{n}\right) = x $$
Wolfram Alpha provides a power series expansion formula which justifies this, but why can't we say the following:
As $n$ is getting bigger and bigger, $\frac{x}{n}$ approaches zero. So $\arctan(x) = 0$, therefore this whole sequence approaches zero as $n$ approaches infinity.
Obviously this is wrong but why it is wrong?
L'Hôpital is overkill in this case. Put $h=x/n$. Then as $x\to\infty$, $h\to0$, and $$n\arctan(x/n)=x\frac{\arctan h}{h}=x\frac{\arctan h-\arctan 0}{h},$$ and the limit of the fraction as $h\to0$ is by definition the derivative of $\arctan x$ at $x=0$.
Try l'Hospital:
$$\lim_{t\to\infty}t\arctan\frac xt=\lim_{t\to\infty}\frac{\arctan\frac xt}{\frac1t}\stackrel{\text{l'H}}=\lim_{t\to\infty}\frac{-\frac x{t^2}\frac1{1+\frac{x^2}{t^2}}}{-\frac1{t^2}}=\lim_{t\to\infty}\,x\frac{t^2}{x^2+t^2}=x$$
and thus the limit is not zero but $\;x\;$
We know that $$\lim_{x \to 0}\frac{\sin x}{x} = 1 = \lim_{x \to 0}\cos x$$ and hence $$\lim_{x \to 0}\frac{\tan x}{x} = 1$$ Putting $\tan x = y$ we see that $x = \arctan y$ and $y \to 0$ as $x \to 0$. Hence we get $$\lim_{y \to 0}\frac{y}{\arctan y} = 1$$ or $$\lim_{y \to 0}\frac{\arctan y}{y} = 1$$ Now we have \begin{align} L &= \lim_{n \to \infty}n\arctan(x/n)\notag\\ &= x\lim_{n \to \infty}\frac{\arctan(x/n)}{x/n}\notag\\ &= x\lim_{y \to 0}\frac{\arctan y}{y} \text{ (putting }y = x/n)\notag\\ &= x\notag \end{align} Coming back to the argument given by OP in his post let's take a simpler example. We know that $1/n \to 0$ as $n \to \infty$ and hence by the argument of OP $n \times (1/n) \to 0$ as $n \to \infty$. But this is clearly not the case. Note that as $n$ takes larger and larger values $1/n$ takes smaller and smaller values, but when we multiply this by $n$ the expression $n \times (1/n)$ becomes $1$. The smallness of $1/n$ is cancelled by the largeness of $n$ when we multiply the two.
In the current question the value of $\arctan (x/n)$ is getting small as $n$ becomes large (check using calculator for $x = 2, n = 1000$). But we have to calculate the limit of $n\arctan(x/n)$ and here again multiplying a small value $\arctan (x/n)$ by a large value $n$ does have a deep impact.
In a formula, you can't replace a part of the formula with its limit without replacing the other part with its own limit. In the present case $n$ tends to $\infty$, while $\arctan \frac xn$ tends to $0$, so this is indeterminate.
A short computation consists in using equivalents: $\arctan u\sim_0 u$, so $$n\arctan\frac xn\sim_\infty n\cdot\frac xn=x,$$ hence the limit is $x$.
take $arctan\frac{x}{n}=\beta $ $$\to tan \beta=\frac{x}{n} \to n=\frac{x}{tan \beta }=x.cot \beta\\ \lim_{n \to \infty} n.arctan \frac{x}{n}= \lim_{\beta \to 0}x.cot \beta .\beta$$
$$ \lim_{\beta \to 0}x.\frac{cos \beta}{sin \beta} .\beta=\\ \lim_{\beta \to 0}\frac{\beta }{sin \beta} .x.cos \beta=\lim_{\beta \to 0}1 .x.cos \beta=\\x.1=\\x$$
