I can't show why x>0 of sin(x)/x is 1!!! I started using $\displaystyle x \leq \sin x \leq x$ then $\displaystyle \frac{x}{x} \leq \frac{\sin x}{x} \leq \frac{x}{x}$ But there is not a single value L for $\displaystyle \lim_{x \to 0} \frac{x}{x}=L$ . Well I have a theory but I'm not sure if it stands. $\displaystyle \lim_{x \to 0^}\frac{x}{x} = (1) = 1 $ $\displaystyle \lim_{x \to 0^+}\frac{x}{x} = 1 $ So $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$ But can I actually do that? I took different kind of limits in both cases. This seems like I'm breaking the inequality since I'm taking different limits at once! It feels like I'm doing something like this: $\displaystyle x + 5 = 5 <=> (x + 5) \cdot 10 = 5 \cdot 20$ I multiplied the two sides of the equation with not the same number. 

The Calculus way to go about this would be L'hopitals rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$, $\frac{\infty}{\infty}$), then $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ So, since $\lim_{x \to 0} \frac{\sin{x}}{x} = \frac{0}{0}$ $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0}\frac{\cos{x}}{1} = \cos{0}=1$ But the method Romsek sent is a much more concrete way to go about it. 
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That's absolutely true, which is why I mentioned the other method is much more concrete. But generally, if I were tossed this problem without having to build from the ground up I would use L'hôpital. 
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Also my original post of taking the left handed side and the right handed side limit is wrong write? −limx→0−xx=−(−1)=1 limx→0+xx=1 So limx→0sinxx=1 
When you took $\lim_{x \to 0^{}} \frac{x}{x}$ , the added negative sign breaks it since the fraction is already negative on the left. This creates the squeeze $\displaystyle \frac{x}{x} \leq \frac{\sin{x}}{x} \leq \frac{x}{x}$ Which would imply $\frac{x}{x} = \sin{x}$ , which isn't always true. 
So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out. 1. You absolutely can NOT use L'Hospital's rule to "show" that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. At best this is circular logic, but I would say it's even worse than that. It is a complete misunderstanding of what L'Hospital's rule means. 2. Romsek's link is technically accurate, but I don't think it's helpful. It begins with 2 inequalities which "can be shown". But given these 2 inequalities it's completely trivial to compute this limit. The important part (which also has all of the insight here) is showing those two inequalities. This is typically done by a geometric argument which is difficult to carry out without drawing the right picture. 
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I don't suppose it would be too much to ask that you post what you consider an adequate solution to the problem being asked would it? 
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