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It is well known that $$\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$$ I know several proofs of this: the geometric proof shows that $\cos(\theta)\leq\frac {\sin(\theta)}{\theta}\leq1$ and using the Squeeze Theorem I conclude that $\lim_{x \rightarrow 0} \frac {\sin(x)}{x} = 1$, other proof uses the Maclaurin series of $\sin(x)$. My question is: is there a demonstration of this limit using the epsilon-delta definition of limit? |
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Define $\sin(x) := x - x^{3}/3! + x^{5}/5! - \cdots$ and show that this is an analytic function and see that we can take derivative term by term so that $\sin'(x) = 1 + x \cdot f(x)$ for some continuous $f$. We have the required limit $= \sin'(0) = 1$. This does not involve any geometric argument and you can trace all the process until you meet $\epsilon$ and $\delta$. ALSO: See Walter Rudin's Principles of Mathematical Analysis Theorem 8.1. |
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For every $x \ne 0$ we have $$ \Big|1-\frac{\sin x}{x}\Big|=\Big|1-\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k+1)!}\Big|\le \sum_{k=1}^\infty\frac{|x|^{2k}}{(2k+1)!}\le\frac13\sum_{k=1}^\infty\frac{|x|^{2k}}{(2k)!}=\frac{\cosh|x|-1}{3}. $$ Given $\varepsilon>0$, let $\delta=\cosh^{-1}(1+3\varepsilon)$. Then $$ 0<|x|\le\delta \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{\cosh|x|-1}{3}\le \varepsilon. $$ Another approach is to notice that $$ x-\sin x\le \frac{x^2}{2} \quad \forall\ x \in [0,\pi]. $$ Since $\sin$ is odd we have $$ -x+\sin x\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,0]. $$ Hence $$ |x-\sin x|\le \frac{x^2}{2} \quad \forall\ x \in [-\pi,\pi]. $$ Given $\varepsilon>0$, we have $$ 0<|x|\le 2\varepsilon \Longrightarrow \Big|1-\frac{\sin x}{x}\Big|\le\frac{|x|}{2} \le \varepsilon. $$ |
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Taking as the definition per the OP's comment, $$\sin x = \sum_{n=0}^\infty {(-1)^n x^{2n+1} \over (2n+1)!}$$ we have that $${\sin x \over x} = \sum_{n=0}^\infty {(-1)^n x^{2n} \over (2n+1)!} =1 - {x^2 \over 6} + {x^4 \over 120} - \cdots \,,$$ which is an alternating series for all $x$ and with decreasing terms for, say, $|x| < 1$. Ok, so let $\epsilon>0$ be arbitrary. Let $\delta = {\rm min}\{\sqrt{6\epsilon},1\}$. Assume $|x| < \delta$. Then $$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}$$ since the series is alternating with decreasing terms, and therefore $$\left\vert 1 - {\sin x \over x}\right\vert \le {x^2 \over 6}< {\big(\sqrt{6\epsilon}\big)^2 \over 6}=\epsilon$$ Thus the limit is $1$. |
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I believe you are looking for a a proof without differentiation but only metric spaces. Define $e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$, define $sin(z) = \frac{1}{2i}(e^{iz}-e^{-iz})$ $\frac{\sin(z)}{z} = 1 +\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}$ Now we prove the sum of the latter term goes to 0: $0\le|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{(2k+1)!}|\le \sum_{k =1 }^\infty |\frac{z^{2k}}{6^{2k}}|=\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}$ As $z\rightarrow 0$, we can pick $\displaystyle N\in\mathbb{N}.\;\forall n\ge N.\;|z_n|<6\;\Longrightarrow |\frac{z_n}{6}|<1$ $\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}} \sum_{k =1 }^\infty |\frac{z}{6}|^{2k} = \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}(\frac{1}{1-|\frac{z}{6}|}-1) = 1-1 =0$ Here we are comparing it with geometric sum. When its absolute value is sandwiched between $0$, the term has to go to $0$. $\Longrightarrow$ $0\le\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$$|\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!}|$ $\le$ $\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}$ $\sum_{k =1 }^\infty |\frac{z}{6}|^{2k}=0$ $\Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 0$ $\Longrightarrow \lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\frac{\sin(z)}{z} = 1 +\lim_{z\rightarrow 0, z\in\mathbb{C}\backslash\{0\}}\sum_{k =1 }^\infty (-1)^k \frac{z^{2k}}{(2k+1)!} = 1$ |
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Here is a more direct answer for this: Since $\cos\theta<\frac{\sin\theta}{\theta}<1$, one can get $$\big|\frac{\sin\theta}{\theta}-1\big|<1-\cos\theta.$$ But $1-\cos\theta=2\sin^2\frac{\theta}{2}\le\frac{\theta^2}{2}$ and hence $$\big|\frac{\sin\theta}{\theta}-1\big|\le\frac{\theta^2}{2}.$$ Now it is easy to use $\varepsilon-\delta$ definition to get the answer. |
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