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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.

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    $\begingroup$ Have you tried using continuity of the $\sin$ function? If you have not proven that for continuous functions: $$\lim_{ x\to a} f(x) = f\left( \lim_{x \to a} x \right)$$ then how about using the sum of angles formula for $\sin$ and see if that simplifies anything. $\endgroup$ – InterstellarProbe Jul 5 '18 at 13:36
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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: $$ \lim_{n \to \infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)} = \lim_{h \to 0}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)} \stackrel{\text{(H)}}{=} \lim_{h \to 0} \frac{\cos(x+h)}{\cos(x+h)} = 1 \, . $$

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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 $\sin x =0$

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

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  • $\begingroup$ I disagree for $x=0$. try to apply L'Hopital's... $\endgroup$ – Jneven Jul 5 '18 at 13:54
  • $\begingroup$ for x=0 we obtain directly the result, we don’t need l’Hopital $\endgroup$ – gimusi Jul 5 '18 at 14:03
  • $\begingroup$ @gimusi Hey, I have a question that is more or less relevant to this question. Do you know why the substitution $h=\frac{1}{n}$ and using the derivative definition fails to give a good answer in this problem? $\endgroup$ – Sorfosh Jul 5 '18 at 20:17
  • $\begingroup$ @Sorfosh If we consider $$\lim_{h\to 0}\frac{\sin\left(a+h\right)-\sin\left(a\right)}{\sin\left(a+h\right)}$$ for $\sin a \neq 0$ it is equal to $0$ and for $\sin a = 0$ it is equal to $1$. In both case there are not the conditions to apply l'Hopital. $\endgroup$ – gimusi Jul 5 '18 at 20:21
  • $\begingroup$ @gimusi I mean $\lim_{n\rightarrow\infty}\frac{\sin\left(x+\frac{1}{n}\right)-\sin\left(x\right)}{\sin\left(x+\frac{1}{n}\right)}=lim_{h\rightarrow0^{+}}\frac{\sin\left(x+h\right)-\sin\left(x\right)}{\sin\left(x+h\right)}=lim_{h\rightarrow0^{+}}\frac{h(sin\left(x+h\right)-\sin\left(x\right))}{h\sin\left(x+h\right)}$ Now split it up and use the derivative definition. $\endgroup$ – Sorfosh Jul 5 '18 at 20:25
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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 $1$.

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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

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

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