GRE (4)

Considering the Maclaurin series of the exponential and cosine functions, we have that

$e^{x^{2}} = 1 + x^{2} + \dfrac{x^{4}}{2!} + O(x^{6}) \Longrightarrow e^{x^{2}} - x^{2} - 1 = \dfrac{x^{4}}{2} + O(x^{6})$ and

$\cos(x) = 1 - \dfrac{x^{2}}{2} + O(x^{4}) \Longrightarrow \cos(x) - 1 = -\dfrac{x^{2}}{2} + O(x^{4})$ .

The given series can thus be written as

$\displaystyle \lim_{x \to 0} \dfrac{\left(\dfrac{x^{4}}{2} + O(x^{6})\right)\left(-\dfrac{x^{2}}{2} + O(x^{4})\right)}{x^{k}} = \lim_{x \to 0} \dfrac{-\dfrac{x^{6}}{4} + O(x^{8})}{x^{k}} = \lim_{x \to 0} \left(-\dfrac{x^{6-k}}{4} + O(x^{8 - k})\right)$ .

For $k \lt 6$ the limit will be $0$ and for $k \gt 6$ the limit will be undefined, (as $6 - k$ will be $\lt 0$ ). Only for $k = \boxed{6}$ will the limit be finite and non-zero, with $L = \boxed{-\dfrac{1}{4}}$ .