# 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.

• 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. Jul 5 '18 at 13:36

## 4 Answers

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

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 $\sin x =0$

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

• I disagree for $x=0$. try to apply L'Hopital's... Jul 5 '18 at 13:54
• for x=0 we obtain directly the result, we don’t need l’Hopital
– user
Jul 5 '18 at 14:03
• @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? Jul 5 '18 at 20:17
• @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.
– user
Jul 5 '18 at 20:21
• @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. Jul 5 '18 at 20:25

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 $1$.

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

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