find: $\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$ 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. 
 A: 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:
$$
\lim_{n \to \infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} =
\lim_{h \to 0}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)} \stackrel{\text{(H)}}{=} \lim_{h \to 0}
\frac{\cos(x+h)}{\cos(x+h)} = 1 \, .
$$
A: From here
$$\lim_{n\longrightarrow\infty} \ \ {1}-\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=1-\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
we have that for $\sin x\neq 0$
$$\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\frac{\sin x}{\sin x}=1$$
and for $\sin x =0$
$$\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=0$$
A: Say $x\ne \pi\cdot k$ where $k\in \mathbb{Z}$ 
$$ \lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\frac{1}{n}}\cdot \lim_{n\rightarrow\infty}{{1\over n}\over{\sin\left(x+\frac{1}{n}\right)} }  = \cos x\cdot 0$$
And if $x= \pi\cdot k$ then the limit is $1$.
A: use $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$
$$\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
= $$\frac{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n}) - \sin(x)}{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n})}$$
for sin(x) not zero
$$\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}$$
$$=\lim_{n\rightarrow\infty}\frac{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n}) - \sin(x)}{\sin(x)\cos(\frac{1}{n}) + \cos(x)\sin(\frac{1}{n})}$$
$$=\frac{\sin(x)\cos(0) + \cos(x)\sin(0) - \sin(x)}{\sin(x)\cos(0) + \cos(x)\sin(0)}$$
$$=\frac{\sin(x).1 + \cos(x).0 - \sin(x)}{\sin(x).1 + \cos(x).0}$$
$$=\frac{\sin(x) - \sin(x)}{\sin(x)}$$
$$ = 0$$
for sin(x) = 0, you can again plug in values and simplify to 
$$\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = 1$$
