Prove a limit using inequalities

[Hernaner28]

Using the inequalities: \sin x \leq x \leq \tan x valid in a zero range, prove that:

\displaystyle\lim_{x \to{0}}{\frac{x}{\sin x}}= 1

Thank you!

 

[Bearded Man]

I'm rather unsure about this, but I'm deeply interested in the answer. I have attempted a solution that may or may not be correct.

What we're trying to prove is that for every epsilon > 0, there is some delta > 0 such that for all x, |x- 0| < delta and |x/sinx - 1| < epsilon. Because of this, we can assign a delta a particular value in terms of the epsilon.

if sinx ≤ x ≤ tanx, then (1/sinx) ≥ (1/x) ≥ (1/tanx)

Factor |(x/sinx) - 1| into |x| * |(1/sinx - 1/x)|

|(1/sinx - 1/x)| ≤ |1/sinx| + |1/x| < |1/sinx| + |1/sinx| by a manipulation of our given inequality = |2/sinx|

Let δ = min(1, ε/|2/sinx|)
from |x| < δ
we have |x| < ε/|2/sinx|
|x|*|(1/sinx + 1/x)| < ε/|2/sinx|*|(1/sinx + 1/x)|
|x/sinx - 1| < ε/|2/sinx|* |2/sinx| by inequality above
|x/sinx - 1| < ε