Evaluating the limit of $x^2 (1-\cos\frac{1}{x})$ when $x$ approaches infinity

Question

I wanted to evaluate the limit $$\lim_{x\to\infty}x^2(1-\cos\frac{1}{x})$$

Since we know that $-1\leq \cos x\leq1$ and that $-1\leq \cos\frac{1}{x} \leq 1$, so by algebraic manipulation, $0\leq x^2(1-\cos\frac{1}{x})\leq2x^2 $.

Why does squeeze theorem fails to obtain a numerical limit here since the actual limit is $\frac{1}{2}$? I know the whole limit would be 0 when $x$ approaches 0.

Answer 1

Following up on my comment, one way to do it without Taylor expansions is so to note that $$ \cos 2y = 1-2\sin^2 y \quad \Leftrightarrow \quad 2\sin^2 \frac{u}{2} = {1-\cos u},$$ so \begin{align} \lim_{x\to \infty} x^2\left(1-\cos\frac{1}{x}\right) &=[u = 1/x] = \lim_{u \to 0^+} \frac{1-\cos u}{u^2}\\ &= \lim_{u\to 0 ^+} \frac{2\sin^2 \frac{u}{2}}{u^2} = \frac{1}{2}\lim_{u \to 0^+} \frac{\sin^2 (u/2)}{(u/2)^2}\\ &= \frac{1}{2} \left(\lim_{u \to 0^+} \frac{\sin(u/2)}{u/2}\right)^2 = \frac{1}{2}\cdot 1^2 = \frac{1}{2}. \end{align}

Answer 2

$\cos(t) = 1 - \frac12 t^2 + O(t^4)$ for $t$ close to $0$ and, so when $|x|$ is large also for $\cos(\frac1x)$.

So substituting $t = \frac{1}{x}$ we have that we consider the limit $\frac{1}{t^2}(1 - \cos(t))$ for $t \downarrow 0$. Using the above estimate for $\cos(t)$, the $1$'s cancel and we are left with $\frac{1}{t^2}(\frac12t^2 + O(t^4)) = \frac12 + O(t^2)$ So indeed the limit is $\frac12$ and series expansion is the easiest (IMHO) way to see it.

Answer 3

This limit is fairly standard in high school: $\;\lim\limits_{u\to 0}\dfrac{1-\cos u}{u^2}=\dfrac12$, so $$x^2\Bigl(1-\cos\frac{1}{x}\Bigr)=\frac{1-\cos\cfrac{1}{x}}{\cfrac1{x^2}}\to\frac12 \;\text{ as }\;\frac1x\to 0.$$