Does an algebraic way to show the limit as x approaches zero of the function (sin x)/x=1 exist?
Since sin x is a transcendental function, there is no "algebraic" proof -- you can prove this using a Taylor series which invoves an infinite number of rational functions, but this is typically reserved for a second semester calculus course.
Here is the typical non-graphical proof: We employ the squeeze theorem -- Let `f(x) <= g(x) <= h(x)` for all x in some interval (c,d), except possibly at the point `a in (c,d)`, and that the limit as `x -> a` of f(x) = lim `x -> a` h(x) = L for some number L. Then `lim x-> a g(x)=L` . Here we are not interested in what happens when x=0.
For `x>0` we know that `sin x < x < tan x`. (See reference blogspot.com) Then divide through by sin x to get `1<x/(sin x)<1/(cos x)` . Now 1/(cos x) =1 at zero and is continuous at zero, so as `x -> 0` from the right the left side and the right side of the compound inequality approach 1. Applying the Squeeze Theorem, letting `f(x)=1,g(x)=x/(sin x),h(x)=1/(cos x)` , and noting that if x/(sin x) approaches 1 so does (sinx)/x, the result holds for x>0.
The same result holds for x<0 as sinx and tanx are odd functions (the inequalities just reverse).
Thus the left-hand and right-hand limits agree, and `lim x->0 (sin x)/x =1`