Does an algebraic way to show the limit as x approaches zero of the function (sin x)/x equals 1 exist?The answer of 1 is obvious from the graph,...

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`