The cancellation of the quadratic terms makes the fourth-order terms decisive. The Taylor expansions give

1 + x^{2} - e^{x^{2}} = - \frac{x ^{4}}{2} - \frac{x ^{6}}{6} + O (x^{8}), x sin x - x^{2} = - \frac{x ^{4}}{6} + \frac{x ^{6}}{120} + O (x^{8}) .

Canceling

x^{4}

for

x \neq = 0

and taking the limit yields

x \to 0 lim \frac{1 + x ^{2} - e ^{x^{2}}}{x sin x - x ^{2}} = 3.

b

Solution

On bounded intervals the exponential series converges uniformly, allowing termwise integration. Since

e^{- t^{2}} = 1 - t^{2} + t^{4} /2 + O (t^{6})

I (x) = \frac{2}{π} (x - \frac{x ^{3}}{3} + \frac{x ^{5}}{10} + O (x^{7})) .

The requested first two nonzero terms are therefore

I (x) = \frac{2 x}{π} - \frac{2 x ^{3}}{3 π} + O (x^{5}) .

I (0) = 0

2 e^{- x^{2}} / π

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