Limits, flipping approaching direction

[plutonium]

I recently found the proof that lim x-> 0 sinx/x = 1
http://www.geocities.com/pkving4math2tor4/4_the_elem_transc_func/4_01_01_03_lim_of_trig_func.htm

There is this one step I don't get.
http://www.geocities.com/pkving4mat..._01_01_03_lim_of_trig_func_files/image020.gif

How did they flip the direction that x approaches from?

 

[courtrigrad]

They let $$t = -x$$

 

[tacman]

Thank you for sharing your discovery of the proof for lim x-> 0 sinx/x = 1. This is a fundamental limit in calculus and it's great that you were able to find the proof for it.

Regarding the step you mentioned, it is common in mathematics to approach limits from both sides. In this case, we are looking for the limit as x approaches 0, but we can approach 0 from the positive side (x>0) and the negative side (x<0). In the proof, they first approach 0 from the positive side and then flip the direction to approach from the negative side. This is a valid approach because the limit should be the same regardless of the direction we approach from.

I hope this helps clarify the step for you. Keep up the great work in exploring and understanding mathematical concepts.