Why does $\frac{|\sin\theta|}{2}<\frac{|\theta|}{2}<\frac{|\tan\theta|}{2}$ not imply that $1>\lim_{\theta\to 0}\frac{\sin\theta}{\theta}>1$?

Good first question! Even if $f(x)<M$ for some value $M$ and for every $x\neq x_0$ in the domain of $f$, you still can't conclude that $\lim\limits_{x\to x_0}f(x)<M$. The most you can say is that $\lim\limits_{x\to x_0}f(x)\leq M$. For example, if $f(x)=1-x^2$, then $f(x)<1$ for all $x\neq0$, but $\lim\limits_{x\to0}1-x^2=1$.

In your case, you have: $$\cos\theta<\frac{\sin\theta}{\theta}<1$$ for every $\theta\neq0$ (at least in a neighborhood of $\theta=0$), so in the limit you have: $$1\leq\lim_{\theta\to0}\frac{\sin\theta}{\theta}\leq 1$$ which is true! So the proof is not broken after all.

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Trigonometry

Limits

Proof Explanation

Solution Verification

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