How come $\frac{1}{\cos x} = 1 + \frac{x^{2}}{2} + o(x^{2})$ as $x \to 0$?

Question

Since $$\cos x = 1 - \frac{x^{2}}{2} + o(x^{3})$$ as $x \to 0,$ we have $$\frac{1}{\cos x} = \frac{1}{1-\frac{x^{2}}{2} + o(x^{3})} = 1 + \frac{x^{2}}{2} + o(x^{3}) + o(\frac{-x^{2}}{2} + o(x^{3})).$$ But I do not see how to write the term on the extreme right as $o(x^{2})$.

Answer 1

You may recall that $$ \frac{1}{1-u}=1+u+O(u^2), \quad u \to 0. $$ Thus $$ \frac{1}{\cos x} = \frac{1}{1-\frac{x^{2}}{2} + o(x^{3})} = 1 + \frac{x^{2}}{2} + o(x^{3}) + O(\frac{-x^{2}}{2} + o(x^{3}))^2 $$ then use $$ O(\frac{-x^{2}}{2} + o(x^{3}))^2 =o(x^{3}). $$