Question 1182300
Using the fact that lim x-->0 (sinx/x)=1 to find lim x-->0 ((1-cosx)/xsinx)) 
<pre>
{{{matrix(2,1,lim,"x->0" )}}}{{{((1-cos(x))/(x*sin(x)))}}}{{{""=""}}}

That given limit is 1 and we can multiply anything by 1 without changing its
value

{{{matrix(2,1,lim,"x->0" )}}}{{{((1-cos(x))/(x*sin(x)))}}}{{{""*""}}}{{{matrix(2,1,lim,"x->0" )}}}{{{(sin(x)/x)}}}{{{""=""}}}

The product of limits equals the limit of products as long as they approach
the same number.

{{{matrix(2,1,lim,"x->0" )}}}{{{(((1-cos(x))/(x*sin(x)))(sin(x)/x)^"")}}}{{{""=""}}}

{{{matrix(2,1,lim,"x->0" )}}}{{{(((1-cos(x))/(x*cross(sin(x))))(cross(sin(x))/x)^"")}}}{{{""=""}}}

{{{matrix(2,1,lim,"x->0" )}}}{{{((1^""-cos(x))/x^2)}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),(((1^""-cos(x))/x^2)((1^""+cos(x))/(1^""+cos(x)))^""))}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),((1-cos^2(x))/(x^2(1^""+cos(x)))))}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),((1-cos^2(x))/(x^2(1^""+cos(x)))))}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),((sin^2(x))/(x^2(1^""+cos(x)))))}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),((sin^2(x)/x^2)^""(1/(1+cos(x))))))}}}{{{""=""}}}

{{{matrix(1,2,matrix(2,1,lim,"x->0" ),((sin(x)/x)^2(1/(1+cos(x))))))}}}{{{""=""}}}

The limit of a product equals the product of limits.

{{{(matrix(1,2,matrix(2,1,lim,"x->0"),(sin(x)/x)^2)^"")}}}{{{""*""}}}{{{(matrix(1,2,matrix(2,1,lim,"x->0"),(1/(1+cos(x)))^""))}}}{{{""=""}}}

In the left parentheses, the limit of a square is a square of the limit.
On the right parentheses, we can substitute what x approaches as long as it
gives a defined value, (not 0/0):

{{{(matrix(1,2,matrix(2,1,lim,"x->0"),(sin(x)/x))^"")^2}}}{{{""*""}}}{{{1/(1+cos(0))}}}{{{""=""}}}

{{{(1^"")^2}}}{{{""*""}}}{{{1/(1+1)}}}{{{""=""}}}

{{{1}}}{{{""*""}}}{{{1/2}}}{{{""=""}}}

{{{1/2}}}      <-- answer

Edwin</pre>