Finding the limit of $\dfrac{\sin x}{2x}$ as $x\to 0$ If $\sin 0= 0$ and $2\cdot 0=0$, how can the answer be $1/2$?
Am I missing a step? 
 A: Well you need to evaluate $$\lim_{x\to 0} \frac{\sin x}{2x}$$ We have
\begin{align}
&\lim_{x\to 0} \frac{\sin x}{2x}\\
&=\frac{1}{2} \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \tag{*}\\
&=\frac{1}{2} \cdot 1\\
&=\frac{1}{2}
\end{align}
To evaluate the limit inside the parenthesis in (*) we are using the well known
theorem from Calculus, that $\lim_{x\to 0} \frac{\sin x}{x}=1$. For a nice geometric proof of this fact you can see Robjohn's answer here. If the OP is taking a calculus I class in the US it is very likely that your textbook uses a similar technique to prove this theorem. This is one of the so called fundamental limits that one has to know in Calculus I. 
Of course once that theorem is proved it can be used freely in evaluating such limits. The point of these kinds of problems is precisely to get you to use that theorem. I hope this helps.
A: $$\lim_{x \to 0} \frac{\sin x }{2x}=\frac{1}{2} \underbrace{\lim_{x \to 0} \frac{\sin x}{x}}_{1}=\frac{1}{2}$$

$\lim_{x \to 0} \frac{\sin x}{x} \overset{ l'Hospital}{=} \lim_{x \to 0} \cos x=1$

A: Remember, we're not evaluating $\frac{\sin(x)}{2x}$ when $x=0$.  If we were to do so, the answer would be undefined (since division by 0 is undefined).  Instead, we are asking what is the limit of the value $\frac{\sin(x)}{2x}$ as $x$ gets closer and closer to 0.  And that's a very different question.
This value can be determined using L'Hopital's rule.
A: The most important reason for introducing the concept of limits in the first place is to deal with limits of expressions in which the numerator and denominator both approach $0$.
That is what happens in the expression $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)} h$.
So it would be a serious mistake to think that if the numerator and denominator both have $0$ as their limits, then the limit of the fraction does not exist.
A: $\sin x\sim x$ around $0$, then $$\lim_{x\to 0}\frac{\sin x}{2x}=\lim_{x\to 0}\frac{x}{2x}=\frac{1}{2}.$$
