Answer to Question #188782 in Real Analysis for TUHIN SUBHRA DAS

Question #188782

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


1
Expert's answer
2021-05-07T11:44:54-0400

"\\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}."


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