First, plug in #x=0# to the function to see what happens.
#color(white)=(sin(0)-0)/(0*ln(1-0^2))#
#=(0-0)/(0*ln1)#
#=0/0#
Since we got #0/0#, we can use L'Hospital's Rule and take the derivative of the top and the bottom of the fraction:
#color(white)=lim_(xrarr0) (sin(x)-x)/(x*ln(1-x^2))#
#=lim_(xrarr0) (d/dx[sin(x)-x])/(d/dx[x*ln(1-x^2)])#
#=lim_(xrarr0) (d/dx[sin(x)]-d/dx[x])/(d/dx[x*ln(1-x^2)])#
#=lim_(xrarr0) (cosx-1)/(d/dx[x*ln(1-x^2)])#
Product rule:
#=lim_(xrarr0) (cosx-1)/(d/dx[x]*ln(1-x^2)+x*d/dx[ln(1-x^2)])#
#=lim_(xrarr0) (cosx-1)/(1*ln(1-x^2)+x*d/dx[ln(1-x^2)])#
#=lim_(xrarr0) (cosx-1)/(ln(1-x^2)+x*d/dx[ln(1-x^2)])#
Chain rule:
#=lim_(xrarr0) (cosx-1)/(ln(1-x^2)+x*1/(1-x^2)*d/dx[1-x^2])#
#=lim_(xrarr0) (cosx-1)/(ln(1-x^2)+x*1/(1-x^2)*-2x)#
#=lim_(xrarr0) (cosx-1)/(ln(1-x^2)-(2x^2)/(1-x^2))#
Plugging in #x=0# still gives #0/0#, so we use L'Hospital's Rule again:
#=lim_(xrarr0) (d/dx[cosx-1])/(d/dx[ln(1-x^2)-(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (d/dx[cosx]-d/dx[1])/(d/dx[ln(1-x^2)-(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (-sinx-0)/(d/dx[ln(1-x^2)-(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (-sinx)/(d/dx[ln(1-x^2)]-d/dx[(2x^2)/(1-x^2)])#
Chain rule:
#=lim_(xrarr0) (-sinx)/(1/(1-x^2)*d/dx[1-x^2]-d/dx[(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (-sinx)/(1/(1-x^2)*-2x-d/dx[(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (-sinx)/((-2x)/(1-x^2)-d/dx[(2x^2)/(1-x^2)])#
#=lim_(xrarr0) (sinx)/((2x)/(1-x^2)+d/dx[(2x^2)/(1-x^2)])#
Quotient rule:
#=lim_(xrarr0) (sinx)/((2x)/(1-x^2)+(d/dx[2x^2]*(1-x^2)-2x^2*d/dx[1-x^2])/(1-x^2)^2)#
#=lim_(xrarr0) (sinx)/((2x)/(1-x^2)+(4x*(1-x^2)-2x^2*-2x)/(1-x^2)^2)#
#=lim_(xrarr0) (sinx)/((2x)/(1-x^2)+(4x-4x^3+4x^3)/(1-x^2)^2)#
#=lim_(xrarr0) (sinx)/((2x)/(1-x^2)+(4x)/(1-x^2)^2)#
#=lim_(xrarr0) (sinx)/((2x(1-x^2))/(1-x^2)^2+(4x)/(1-x^2)^2)#
#=lim_(xrarr0) (sinx)/((2x(1-x^2)+4x)/(1-x^2)^2)#
#=lim_(xrarr0) (sinx*(1-x^2)^2)/(2x(1-x^2)+4x)#
Plugging in #x=0# STILL gives #0/0#, so we use L'Hospital's Rule for a final time:
#=lim_(xrarr0) (d/dx[sinx*(1-x^2)^2])/(d/dx[2x(1-x^2)+4x])#
Product rule:
#=lim_(xrarr0) (d/dx[sinx]*(1-x^2)+sinx*d/dx[1-x^2])/(d/dx[2x(1-x^2)+4x])#
#=lim_(xrarr0) (cosx*(1-x^2)+sinx*-2x)/(d/dx[2x(1-x^2)+4x])#
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(d/dx[2x(1-x^2)]+d/dx[4x])#
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(d/dx[2x(1-x^2)]+4)#
Product rule:
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(d/dx[2x]*(1-x^2)+2x*d/dx[1-x^2]+4)#
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(2*(1-x^2)+2x*-2x+4)#
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(2-2x^2-2x^2+4)#
#=lim_(xrarr0) (cosx-x^2cosx-2xsinx)/(6-4x^2)#
Finally, plug in #x=0# to get the result:
#=(cos0-0^2cos0-2*0sinx)/(6-4*0^2)#
#=(cos0)/(6)#
#=1/6#
That's the limit. Hope this helped! (Sorry for such a long answer)
