Demonstrating Limit Equalities for x→0

[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?

 

[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.

 

[tmc]

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

 

[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.

 

[masudr]

Or try the Taylor expansions about x=0.

 

[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.

 

[shaner-baner]

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