$$\lim_{x \to 0} \frac{e^{x^2} - \cos x}{\sin^2 x}$$ is equal to

We evaluate $$\displaystyle\lim_{x \to 0} \frac{e^{x^2} - \cos x}{\sin^2 x}$$.

Since direct substitution gives the indeterminate form $$\dfrac{0}{0}$$, we use Maclaurin series. Recall $$e^{x^2} = 1 + x^2 + \dfrac{x^4}{2} + \cdots$$, $$\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$$, and $$\sin^2 x = x^2 - \dfrac{x^4}{3} + \cdots$$.

The numerator becomes $$e^{x^2} - \cos x = \left(1 + x^2 + \dfrac{x^4}{2} + \cdots\right) - \left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots\right) = \dfrac{3x^2}{2} + \dfrac{11x^4}{24} + \cdots$$.

The denominator is $$\sin^2 x = x^2 - \dfrac{x^4}{3} + \cdots$$.

Dividing: $$\frac{e^{x^2} - \cos x}{\sin^2 x} = \frac{\dfrac{3x^2}{2} + O(x^4)}{x^2 + O(x^4)} = \frac{\dfrac{3}{2} + O(x^2)}{1 + O(x^2)}$$.

As $$x \to 0$$, this tends to $$\dfrac{3}{2}$$.