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

Question

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 1

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

Answer 2

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 $\sin x =0$

$$\lim_{n\longrightarrow\infty}\frac{\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=0$$

Answer 3

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 $1$.

Answer 4

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

$$\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = 1$$