Differentiation or integration?

The trick here is to apply L'Hôpital's rule

Apply Fundamental Theorem of Calculus part 1

$\large \begin{aligned} \displaystyle \lim_{x \to 0 } \frac {x^2 - \int_0^{x^2} e^{-y^2} \mathrm{d} y }{x^6 } & = & \lim_{x \to 0 } \frac {2x - 2x e^{-x^4}}{6x^5} \\ & = & \lim_{x \to 0 } \frac {1}{3} \left ( \frac {1 - e^{-x^4}}{x^4} \right ) \\ & = & \lim_{y \to 0 } \frac {1}{3} \left ( \frac {1 - e^{-y}}{y} \right ) \\ & = & \lim_{y \to 0 } \frac {1}{3} e^{-y} \\ & = & \frac {1}{3} \cdot e^{-0} = \frac {1}{3} \\ \end{aligned}$

Hence, $a = 1, b= 3, \Rightarrow a+b= \boxed {4}$

Take LCM , write as $L = \displaystyle \lim_{x \to 0} \frac{x^2 - \int_{0}^{x^2} e^{-x^2} \text{dx}}{x^6}$

Apply LH rule to get $L = \displaystyle \lim_{x \to 0} \frac{2x - e^{-(x^2)^2} \times 2x}{6x^5}$

= $\displaystyle \lim_{x \to 0} \frac{2x(1 - e^{-x^4})}{6x^5}$ ,

= $\displaystyle \lim_{x \to 0} \frac{2x(x^4)}{6x^5} = \boxed{\frac{1}{3}}$ (By expanding $e^{-x^4} = 1 - x^4 + \frac{x^8}{2!} \dots$ )

We begin by adding the two expressions to get:

$\lim_{x \to 0}\frac{x^{2}-\int_0^{x^2} \mathrm{e}^{-x^{2}}\,\mathrm{d}x}{x^{6}}$

This limit evaluates to $\frac{0}{0}$ , so we can use l'Hôpital's rule and differentiate with respect to $x^{2}$ to get:

$\lim_{x \to 0}\frac{1-e^{-(x^{2})^{2}}}{3(x^{2})^{2}}$

$= \lim_{x \to 0}\frac{1-e^{-x^{4}}}{3x^{4}}$

This limit still evaluates to $\frac{0}{0}$ , so we can again use l'Hôpital's rule. However, this time we will differentiate with respect to $x^{4}$ to get:

$\lim_{x \to 0}\frac{e^{-x^{4}}}{3}$

$= \boxed{\frac{1}{3}}$