We are asked to evaluate the limit: $\lim_{x \to +\infty} x^2 (1 - \cos(\frac{1}{x}))$

1. Problem Description

We are asked to evaluate the limit:

$\lim_{x \to +\infty} x^2 (1 - \cos(\frac{1}{x}))$

2. Solution Steps

Let $u = \frac{1}{x}$. As $x \to +\infty$, we have $u \to 0$. Therefore, the limit can be rewritten as:

$\lim_{u \to 0} \frac{1}{u^2} (1 - \cos(u))$

We can multiply the expression by $\frac{1 + \cos(u)}{1 + \cos(u)}$ to get:

$\lim_{u \to 0} \frac{1}{u^2} \frac{(1 - \cos(u))(1 + \cos(u))}{1 + \cos(u)} = \lim_{u \to 0} \frac{1}{u^2} \frac{1 - \cos^2(u)}{1 + \cos(u)}$

Using the identity $\sin^2(u) + \cos^2(u) = 1$, we have $1 - \cos^2(u) = \sin^2(u)$.

So the limit becomes:

$\lim_{u \to 0} \frac{\sin^2(u)}{u^2} \frac{1}{1 + \cos(u)} = \lim_{u \to 0} \left(\frac{\sin(u)}{u}\right)^2 \frac{1}{1 + \cos(u)}$

We know that $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$ and $\lim_{u \to 0} \cos(u) = 1$. Thus,

$\lim_{u \to 0} \left(\frac{\sin(u)}{u}\right)^2 \frac{1}{1 + \cos(u)} = (1)^2 \frac{1}{1 + 1} = 1 \cdot \frac{1}{2} = \frac{1}{2}$

3. Final Answer

$\frac{1}{2}$