Limit

**Stuck Man** · November 19th 2010, 06:18 AM

Why is the limit 1?

**Plato** · November 19th 2010, 06:30 AM

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

**Stuck Man** · November 19th 2010, 06:30 AM

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

**HallsofIvy** · November 19th 2010, 11:15 AM

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.

**Educated** · November 19th 2010, 12:40 PM

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.

**HallsofIvy** · November 20th 2010, 02:22 AM

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 $\theta= \frac{\delta x}{2}$ .

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