Prove the following statements are false:
$e^x-1=\mathcal{O}(x^2)$ as $x\to 0$
$x^{-2}=\mathcal{O}(\cot x)$ as $x\to 0$
For the first one, I tried to graph them and to me it seems like $e^x-1$ blows up eventually. But I don't know how to prove it formally.
And for the second one, I have no idea. Can anyone give me a hint?
Question 1. Using Taylor series for $e^x$ $$e^x = 1 + x + \frac{x^2}{2}+\frac{x^3}{6}...$$ We have as x approach 0
$$e^x -1 = x + \frac{x^2}{2}+\frac{x^3}{6}...$$
$$e^x -1 = x + O(x^2) \neq O(x^2)$$
Question 2. Using Taylor series for $cot(x)$ $$cot(x) = \frac{1}{x} - \frac{x}{3} - \frac{x^3 }{45} - ...$$ Taking the Big O notation
$$O(cot(x)) = O(\frac{1}{x} - \frac{x}{3} - \frac{x^3 }{45} - ...)$$
$$O(cot(x)) = O(\frac{1}{x})$$ However
$$x^{-2} = O(x^{-2}) \neq O(\frac{1}{x})$$
$$x^{-2} \neq O(cot(x))$$ Look up Taylor series from Wolfram Alpha