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