The limit expression you provided is:

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

Steps to solve the limit:

1. Analyze the behavior of $\cos\left( \frac{1}{x} \right)$ as $x \to \infty$:

   As $x \to \infty$, $\frac{1}{x} \to 0$. We know that: $\cos(0) = 1$ Therefore, $\cos\left( \frac{1}{x} \right) \to 1$ as $x \to \infty$.

2. Use the small angle approximation:

   For very small angles, we can use the approximation: $1 - \cos(\theta) \approx \frac{\theta^2}{2} \quad \text{when} \quad \theta \to 0$ In this case, $\theta = \frac{1}{x}$, so: $1 - \cos\left( \frac{1}{x} \right) \approx \frac{\left(\frac{1}{x}\right)^2}{2} = \frac{1}{2x^2}$

3. Substitute the approximation into the limit expression: $\lim_{{x \to +\infty}} x^2 \left( 1 - \cos\left( \frac{1}{x} \right) \right) \approx \lim_{{x \to +\infty}} x^2 \cdot \frac{1}{2x^2}$

4. Simplify the expression: $\lim_{{x \to +\infty}} \frac{1}{2} = \frac{1}{2}$

Final Answer:

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

Would you like further clarification or details on this?

Here are some related questions to explore:

1. How do small angle approximations work in other trigonometric limits?
2. What happens to the limit if we replace $\cos\left(\frac{1}{x}\right)$ with $\sin\left(\frac{1}{x}\right)$?
3. Can L'Hopital's rule be applied to this limit? If so, how?
4. How would the solution change if the limit was as $x \to 0$ instead of $x \to \infty$?
5. What role does Taylor series expansion play in solving limits involving trigonometric functions?

Tip: For limits involving trigonometric functions as their argument tends to zero, small angle approximations or Taylor series expansions can often simplify calculations.