limits - find: $lim_{nrightarrowinfty}frac{sinleft(x+frac{1}{n}right)-sinleft(xright)}{sinleft(x+frac{1}{n}right)}$

Find:

$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ as $x\in\mathbb{R}$
My progress:

$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\lim_{n\longrightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)}{\sin\left(x+\frac{1}{n}\right)}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=$$

$$\lim_{n\longrightarrow\infty} \ \ {1}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=1-\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$ at this point I got stuck.

I can't evaluate the Taylor series of $\sin(x+\frac{1}{n})$ because $n$ is not fixed

(even if we'll suppose that there exist some $\epsilon>0$ and there exists $N\in\mathbb{N}:\forall n\geq N$ s.t:

$$-\epsilon<\frac{1}{n}<\epsilon$$
it doesn't seem like a formal argument to me)

(I might be very wrong - it's only my intuition).

Also trying to apply L'Hopital's rule for this expression isn't much helpful.

Answer

If $\sin(x) = 0$ then

$$\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = 1 \to 1$$
for $n \to \infty$, otherwise $\sin\left(x+\frac{1}{n}\right) \to \sin(x) \ne 0$ and therefore

$$\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} \to \frac{\sin(x) - \sin(x)}{\sin(x)} = 0 \, .$$

So there is no need to use L'Hospital's rule in the case $\sin(x) = 0$,
but doing so would give the same result:

$$\lim_{n \to \infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = \lim_{h \to 0}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)} \stackrel{\text{(H)}}{=} \lim_{h \to 0} \frac{\cos(x+h)}{\cos(x+h)} = 1 \, .$$