I'd like to see a proof that lim x->0 x/sinx = 1.
It is given that lim x->0 sinx/x = 1
The Attempt at a Solution
I can confirm this limit quite easily by graphing - the curve obviously approaches one, and the left- and right-handed limits are equal. However, assuming that either I can't graph or don't have the time to plot points, (on a timed standardized test that doesn't allow graphing calculators, for example) how would I know this to be true?
I thought about the fact that perhaps some property out there exists saying that if lim x->a f(x)/g(x) = L, then its reciprocal, lim x->a g(x)/f(x) = 1/L. However, I don't believe this to be true, since lim x->-2 (x^2+4x+4)/(x+2) = 0, and the reciprocal lim x->-2 (x+2)/(x^2+4x+4) = DNE
Perhaps this is because the reciprocal of 0 is undefined, and because of this, this is one instance where my property fails. In which case, I can make a simple restriction which allows me to use this property for all other instances. Of course, this isn't very convincing, and leaves me quite skeptical.
Also - I know of L'Hopital's rule, and could easily use that in this instance, but we just learned limits, so that's off the table for now.
If you must use some special techniques to prove this, please do. But I believe if I can show that my property works, then I will have this proof completed. Furthermore, I could use that line of thinking for other limits I run into to save me some time.
Thanks for your help in this matter!