For every $x \in \mathbb{R}$ holds that $ - 1\le \sin(x) \leq 1$. So in my opinion the following inequality is true $ -\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}$ and we have $\lim _{n \rightarrow \infty} \frac{1}{x}$, $\lim _{n \rightarrow \infty} -\frac{1}{x} = 0$. But $\lim _{n \rightarrow \infty} \frac{\sin(x)}{x} \neq 0 $. Where's my fallacy?
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But that limit is zero. Are you thinking of the limit as $x\to 0$? In that case the squeeze theorem won’t work as you’ve set it up. – csch2 Jan 21 '20 at 23:55
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I think you meant $\lim_{x \to \infty} \dfrac{\sin x}{x}$, in which case, yes this limit is $0$. What I believe you are confusing this with is an entirely different limit: $\lim_{x \to 0} \dfrac{\sin x}{x}=1$. Notice one is to infinity, the other is to $0$.
mathematics2x2life
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The inequality you used is just an application of the well-known Squeeze-theorem. For $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ See:How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
PinkyWay
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