Uniform convergence with sine functions

Question

Let $f_n = \sin(\frac{3n}{4n+5}x) $ as $f_n:\mathbb{C} \rightarrow \mathbb{C} $.
I try to determine if $f_n$ is uniformly convergent on the open intervals $I_1=(0; 1)$ and $I_2(1; \infty)$.
Let $\epsilon > 0$ and $n_0>0$.
$f_n$ is uniformly convergent, iff
$$\forall x \in I: |f_n(x)-f(x)|<\epsilon$$
I figured out that $$f(x)=\lim_{n \to \infty}(f_n(x))=\sin(\frac{3}{4}x)$$.
So it follows: $$\forall x \in I: \biggl|\sin\Bigl(\frac{3n}{4n+5}x\Bigr)-\sin\Bigl(\frac{3}{4}x\Bigr)\biggr| < \epsilon $$ But how do I find out about $\epsilon $?
How do I find out, if it's really uniformly convergent?

Answer 1

That sequence converges uniformly to $f$ on $(0,1)$. In fact, since $\sin'=\cos$, and since $|\cos|$ is bounded above by $1$, you have$$(\forall x,y\in\Bbb R):|\sin(x)-\sin(y)|\leqslant|x-y|.$$So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $n\geqslant N\implies\left|\frac{3n}{4n+5}-\frac34\right|<\varepsilon$. And then, if $x\in(0,1)$ and $n\geqslant N$,$$\left|\sin\left(\frac{3n}{4n+5}x\right)-\sin\left(\frac34x\right)\right|\leqslant\left|\left(\frac{3n}{4n+5}-\frac34\right)x\right|<\varepsilon.$$

However, your sequence does not converge uniformly on $(1,\infty)$. Can you see why?