Question

Find the following limit x tends to 0 (1-cosx^2/x^2 - x^2 sin x^2)

Answer 1

$\lim_{x \to 0} \dfrac{1 - \cos (x^2)}{x^2 \sin(x^2)}.$

Expansion of cos x and sinx is

$cos (x^2) = 1 - \frac{x^4}{2!} + o(x^5) ,\\[9pt] sin (x^2) = x^2 + o(x^4)$

herefore, we compute the limit as

$\lim_{x \to 0} \dfrac{1 - \cos (x^2)}{x^2 (\sin (x^2))} \\[9pt]= \lim_{x \to 0} \dfrac{1 - 1 + \dfrac{1}{2}x^4 + o(x^5)}{x^2(x^2+o(x^4))} \\[9pt] = \lim_{x \to 0} \dfrac{\frac{1}{2}x^4 + o(x^5)}{x^4 + o(x^6)} \\[9pt] = \lim_{x \to 0} \dfrac{\frac{1}{2} + \dfrac{o(x^5)}{x^4}}{1 + \dfrac{o(x^6)}{x^4}} \\[9pt] = \frac{1}{2}.$