How do you prove: #lim_(x->0) sin(x)/x = 1# without using l'hopital's rule (or the derivative of sin(x) at all)?
I saw a proof which uses the limit definition of a derivative to prove #d/(dx)sin(x)=cos(x)#
but part of the proof relied upon assuming that:
#lim_(x->0) sin(x)/x = 1# .
It is not shown explicitly in the proof how this limit is evaluated.
The only way I know how to evaluate that limit is using l'hopital's rule which means the derivative of #sin(x)# is already assumed to be #cos(x)# and will obviously lead to some circular logic thereby invalidating the proof.
I was hoping someone might be able to present a method which does not take the derivative of #sin(x)# to find the limit (and is also a bit more algebraically rigorous than sticking numbers in a spreadsheet and seeing they approach one). Thanks!
For context the proof I saw is here:
I saw a proof which uses the limit definition of a derivative to prove
but part of the proof relied upon assuming that:
It is not shown explicitly in the proof how this limit is evaluated.
The only way I know how to evaluate that limit is using l'hopital's rule which means the derivative of
I was hoping someone might be able to present a method which does not take the derivative of
For context the proof I saw is here:
1 Answer
Use the squeeze theorem.
Explanation:
A proof using geometry and the squeeze theorem is here:
https://socratic.org/questions/how-do-you-use-the-squeeze-theorem-to-find-lim-sin-x-x-as-x-approaches-zero
The Squeeze Theorem can be proved directly from the definition of the limit of a function.
Outline:
Suppose (1)
also suppose that (2)
Given
By (2) we can make
#f(x) <= g(x) <= h(x)# and
#L-epsilon < f(x) < L+epsilon# (for#abs(x-c) <# some#delta_1)# )
By (2) we can make also make
#f(x) <= g(x) <= h(x)# and
#L-epsilon < g(x) < L+epsilon# (for#abs(x-c) <# some#delta_2)# )
Now using
So
(OK I was going to type an outline but that's a fairly complete proof.)
