# Simple problem

1. Nov 29, 2005

### Werg22

I need a solid demonstration of those two equalities

$${_\lim{x}{\rightarrow} 0}\frac{\sin {x}}{x} = 1$$

$${_\lim{x}{\rightarrow} 0}\frac{1-\cos {x}}{x} = 0$$

How to do so?

Last edited: Nov 29, 2005
2. Nov 29, 2005

### amcavoy

Use the following fact and do some algebra / trig. to rearrange, then use the Squeeze Theorem:

$$\sin{x}\cos{x}\leq x\leq\tan{x}$$

For the second, multiply the numerator and denominator by 1+cos(x) and use the results of the first when simplifying.

3. Nov 29, 2005

### tmc

is l'hopital's rule not a good enough demonstration?

4. Nov 29, 2005

### amcavoy

Yeah, that works. The only reason I did so otherwise is because generally these limits are introduced at the beginning of an introductory calculus course, before students have had derivatives or L'Hopital's Rule. At least, that is how it was for me.

5. Nov 29, 2005

### masudr

Or try the Taylor expansions about x=0.

6. Nov 29, 2005

### shaner-baner

triangles

It is usually done without calculus. Just because it is used to define the derivatives of certain functions. If you compare the areas in the graph you can come up with the inequalities and use the squeeze theorem. Probably works out to the same formula above, but with a little more motivation.

7. Nov 29, 2005

### shaner-baner

sorry, forgot the picture. Sorry it's not very "pretty"

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