How do you find the limit of #(3 sin (x)(1 - cos(x))) / (x^2) # as x approaches 0?

To find #Lt_(x->0)(3sin(x)(1-cos(x)))/x^2#, let us divide it in two parts

#Lt_(x->0)(3sin(x))/x xx Lt_(x->0)(1-cos(x))/x#

Now as each of these is of the form #0/0# when #x->0#,

we can use L'Hospital's Rule,

and as such #Lt_(x->0)(3sin(x))/x xx Lt_(x->0)(1-cos(x))/x#

= #Lt_(x->0)(d/(dx)3sin(x))/1 xx Lt_(x->0)(d/(dx)(1-cos(x)))/1#

= #Lt_(x->0)(3cos(x))/1 xx Lt_(x->0)(0-(-sin(x)))/1#

= #3xx0/1=0#
graph{(3sin(x)(1-cos(x)))/x^2 [-2.5, 2.5, -1.25, 1.25]}

#1-cos(x)=cos(0)-cos(x)#

we know that

#cos(a+b)-cos(a-b)=-2sin(a)sin(b)#

calling

#{(a+b=0),(a-b=x):}#

and solving for #a,b#

we have #a=x/2,b=-x/2# so

#cos(0)-cos(x)=2sin^2(x/2)#

then

#(3sin(x)(1-cos(x)))/x^2 = (6sin(x)sin^2(x/2))/x^2#

Now using the fundamental result

#lim_(x->0)sin(x)/x=1# we have

#lim_(x->0)(3sin(x)(1-cos(x)))/x^2=6/4lim_(x->0)sin(x)lim_(x->0)(sin(x/2)/(x/2))^2 = 3/2 cdot 0 cdot 1 = 0#

We have

#lim_(xrarr0)sinx/x = 1# #" "# and #" "# #lim_(xrarr0)(1-cosx)/x = 0#.

These limits, together with the product property of limits allow us to write:

#lim_(xrarr0)(3sinx(1-cosx))/x^2 = 3lim_(xrarr0)(sinx/x) lim_(xrarr0) (1-cosx)/x#

# = 3(1)(0) = 0#