Proof of limit as x tends to 0 of sin x/x =1 using the inequality -|x| <= sin x<= |x| I already know the proof using the Unit circle. But I was just wondering if it was possible to prove using the above inequality. I could not do it and my professor said that the inequality was too broad to squeeze. Do ya'll think that it is possible? If so then how to do it? 
 A: By only assuming that squeezing inequality, what you are trying to do is prove the following claim:

Claim. If $f$ is a function that satisfies $-|x| \le f(x) \le |x|$ then $\lim_{x \to 0} \frac{f(x)}{x} = 1$.

The claim is false, although it is not quite enough to just manipulate the expression to $-1 \le \frac{f(x)}{|x|} \le 1$.
To see this is false, consider, e.g. $f(x) = x \sin(1/x)$, and $f(0) = 0$. Then not only does $\lim_{x \to 0} f(x) / x$ fail to exist, but there are sequences $x_n^{\alpha}$ tending to zero for which $f(x_n) \to \alpha$ for any $\alpha \in [-1, 1]$.
A: You're trying to find the limit of $\frac{\sin x}{x}$ but the inequality you give holds for $\sin x$.
If you tried to transform this inequality in terms of $\frac{\sin x}{x}$, the best you can do is divide everything by $x$ to get: $$-\frac{|x|}{x}\le\frac{\sin x}{x}\le \frac{|x|}{x}$$ when $x$ is positive and $$\frac{|x|}{x}\le\frac{\sin x}{x}\le -\frac{|x|}{x}$$ when $x$ is negative.
But unfortunately this all simplifies to $$-1\le \frac{\sin x}{x}\le 1$$
The limits of the LHS and RHS side are not equal, so the squeeze theorem does not apply. That is why arguments like the unit circle argument are used.
