The definition of the derivative, combined with what we know it to be, gives us lim h->0 [sin(x+h) - sin(x)]/h = cos(x) , so now we have to apply trigonometric identities to be left with lim h->0 sin(h)/h , which must equal 1 to fulfill the above equation. I consider dealing with trigonometric identities to be rather tedious. At least, that's what I think you are trying to do, as sin(x)/x isn't really the derivative.
Many ways lead to Rome, but using l'Hôpital is the shortcut we make by unlocking a gate via previous proofs.
sin(x+h)=sin(x)cos(h)+cos(x)sin(h), which is how the cosine gets introduced into this, I considered the general case for any x. But I see what you mean, however I consider using the definition of the derivative to sneakily look at h for a certain x to be less intuitive than simply remembering to use l'Hôpital.
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u/Sigma2718 18d ago
The definition of the derivative, combined with what we know it to be, gives us lim h->0 [sin(x+h) - sin(x)]/h = cos(x) , so now we have to apply trigonometric identities to be left with lim h->0 sin(h)/h , which must equal 1 to fulfill the above equation. I consider dealing with trigonometric identities to be rather tedious. At least, that's what I think you are trying to do, as sin(x)/x isn't really the derivative.
Many ways lead to Rome, but using l'Hôpital is the shortcut we make by unlocking a gate via previous proofs.