First it should be obvious that is a and b are both positive then the limit is 0 so the function is continuous. I would let [tex]u= x^2[/tex] so the limit becomes [tex]\lim_{u\to 0}u^a sin^b(u)[/tex]. Because [tex]\lim_{x\to 0} \frac{sin(x)}{x}= 1[/tex] I would then rewrite it as [tex]\lim_{x\to 0} u^{a- b}\left(\frac{sin(u)}{u}\right)^b[/tex]. It looks to me like [tex]\lim_{x\to 0} \left(\frac{sin(u)}{u}\right)^b= 1[/tex] for any b so we must have [tex]\lim_{u\to 0} u^{a- b}= 0[/tex] and that requires that a> b.