I have this limit:
$$
\lim_{x\to \infty}x^2-x^2\cdot \cos\left(\frac{1}{x}\right)
$$
For me my initial answer would be zero as:
$$
\lim_{x\to \infty}x^2-x^2\cdot \cos\left(\frac{1}{x}\right)=\lim_{x\to \infty}x^2-\lim_{x\to \infty}x^2\cdot\lim_{x\to \infty}\cos\left(\frac{1}{x}\right)
$$
Which is:
$$
\infty-\infty\cdot1=0
$$
But after looking at wolfram alpha and doing a series expansion of $\cos(x)$ i see that the answer is in fact $1/2.$
Why is my original thinking incorrect?
Best Answer
Hint: $\cos(x) = 1-\dfrac{x^2}{2}+O(x^4)$ or $\cos(1/x) = 1 - \dfrac{1}{2x^2}+O\left(\dfrac{1}{x^4}\right)$