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That works. You could also use l'Hopital's rule and get a result quicker
Yes. But this way is simpler for undergraduates.
No, you cannot, because the proof that the derivative of sin(x) is cos(x) requires that the limit as x approaches 0 of sin(x)/x to be 1 which is literally the thing were trying to prove, thats a bad case of circular reasoning
Erweitern sie den sinx durch die Taylor Reihe
You just proved 1 < 1. I don't think so!
it's just approximation
Yes, he wasn't very careful about his inequalities, since if x = 0 those are no longer strict inequalities. In other words, we need to say that sin(x)
Why in heaven's name would I want to.? It adds nothing to our betterment of life.🤮
![Proofs: Lim sinx/x =1 and lim [cosx -1] /x =0 as x goes to zero from geometry](http://i.ytimg.com/vi/W0jiodfVGx0/mqdefault.jpg)
![Proofs: Lim sinx/x =1 and lim [cosx -1] /x =0 as x goes to zero from geometry](/img/tr.png)
That works. You could also use l'Hopital's rule and get a result quicker
Yes. But this way is simpler for undergraduates.
No, you cannot, because the proof that the derivative of sin(x) is cos(x) requires that the limit as x approaches 0 of sin(x)/x to be 1 which is literally the thing were trying to prove, thats a bad case of circular reasoning
Erweitern sie den sinx durch die Taylor Reihe
You just proved 1 < 1. I don't think so!
it's just approximation
Yes, he wasn't very careful about his inequalities, since if x = 0 those are no longer strict inequalities. In other words, we need to say that sin(x)
Why in heaven's name would I want to.? It adds nothing to our betterment of life.🤮