1. ## Limit

Why is the limit 1?

2. Do you understand that $\displaystyle\lim_{t\to 0}\frac{\sin(t)}{t}=1~?$

3. I think I see now that it is 1 when delta x is 0.0001 or some other small value near 0.

4. No, it is close to 1 "when delta x is 0.0001 or some other small value near 0". The crucial point is that it keeps getting closer and closer to 1 as delta x gets closer to 0.

5. You could use used L'hospital's rule to solve the limit.

$\lim \limits_{x \rightarrow 0} \dfrac{f(x)}{g(x)} = \lim \limits_{x \rightarrow 0} \dfrac{f'(x)}{g'(x)}$

$\lim \limits_{\sigma x \rightarrow 0} \dfrac{\sin{\frac{\sigma x}{2}}}{\frac{\sigma x}{2}}$ $= \lim \limits_{\sigma x \rightarrow 0} \dfrac{0.5 \cos \frac{\sigma x}{2}}{0.5}$

Substitute in $\sigma x = 0$ and you will get your answer.

6. Whether that is valid or not depends upon exactly how you have proved that the derivative of sin(x) is cos(x). Most textbooks use the limit $\displaytype\lim_{x\to 0}\frac{sin(\theta)}{\theta}= 1$. If you already have that then the given limit follows immediately by letting $\theta= \frac{\delta x}{2}$.