How would I go about obtaining this limit? $\lim _{x\to \:0}\frac{\int _0^{x^2}\frac{t^3}{1+t^6}dt}{x^8}$
My original plan is to use l'hopital's rule, but I realise i do not have the conditions required (i.e. 0/0)
 A: $$\lim_{x\to 0}\frac{\int _0^{x^2}\frac{t^3}{1+t^6}dt}{x^8}$$
Well, note that you actually do have the required conditions: 
$$\lim_{x\to a}\int_a^xf(t)dt=0$$
So: 
$$\lim_{x\to 0}\frac{\int _0^{x^2}\frac{t^3}{1+t^6}dt}{x^8}=\lim_{x\to0}\frac{\frac{d}{dx}\left[\int _0^{x^2}\frac{t^3}{1+t^6}dt\right]}{8x^7}=\lim_{x\to0}\frac{2x\cdot x^6}{8x^7 \cdot (1+x^{12})}=\lim_{x\to 0}\frac{1}{4(1+x^{12})}=\frac{1}{4(1)}=\frac14$$
A: Both the numerator and denominator approach $0$ when $x\to 0$. Hence, $$\lim _{x\to 0}\frac{\int _0^{x^2}\frac{t^3}{1+t^6}dt}{x^8}=\lim_{x\to 0 }\frac{2x\cdot\left(\frac{x^6}{1+x^{12}}\right)}{8x^7}=\lim_{x\to 0 }\frac{1}{4\left(1+x^{12}\right)}=\frac{1}{4}$$
A: One can also use Squeeze Theorem here. If $0\leq t\leq u$ then $$\frac{t^{3}}{1+u^{6}}\leq\frac{t^{3}}{1+t^{6}}\leq t^{3}$$ and integrating in the interval $[0, u]$ we get $$\frac{u^{4}}{4(1+u^{6})}\leq\int_{0}^{u}\frac{t^{3}}{1+t^{6}}\,dt\leq\frac{u^{4}}{4}$$ and hence usig Squeeze Theorem we get $$\lim_{u\to 0^{+}}\frac{1}{u^{4}}\int_{0}^{u}\frac{t^{3}}{1+t^{6}}\,dt=\frac{1}{4}$$ and this is same as limit in question via the substitution $x^{2}=u$.
