0

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?

Bernard
  • 173,269
  • 10
  • 66
  • 166
Anna Saabel
  • 525
  • 3
  • 13
  • 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

2 Answers2

4

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
  • 12,738
  • 2
  • 30
  • 49
0

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 $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Invisible
  • 4,474
  • 4
  • 12
  • 40