#lim_(x to 0^+)(x^alog(x))/(log(1+x^2)(sin(x^2)-sin^2(x)))#, #a>0#
#=x^a/log(1+x^2)*log(x)/(sin(x^2)-sin^2(x))#
Now let observe #x^a/log(1+x^2)#
The limit is #0/0#, and both functions are continuous on # 0^+#.
Using L'Hôpital's therorem
#=(ax^(a-1))/((2x)/(1+x^2))#
#=a/2(x^(a-2)+x^(a-1))#
And #lim_(x to 0^+) a/2(x^(a-2)+x^(a-1))=0#
Now we have #lim_(x to 0^+) ((a/2(x^(a-2)+x^(a-1)))log(x))/(sin(x^2)-sin^2(x))#
Now Let's take a look at #(a/2(x^(a-2)+x^(a-1)))/(sin(x^2)-sin^2(x))#
Using L'Hôpital's theorem
#=a/2((a-2)x^(a-3)+(a-1)x^(a-2))/(2xcos(x^2)-2cos(x)sin(x))#
Using L'Hôpital again
#f=a/4((a-2)(a-3)x^(a-4)+(a-1)(a-2)x^(a-3))/(cos(x^2)+2x²sin(x²)-sin(2x)) #
With #lim_(x to 0^+)f(x) =0#
Finally, #lim_(x to 0^+)f(x)log(x)=lim_(x to 0^+)log(x^(f(x)))=log(1)=0#
\0/ Here's our answer !
