I have to find this limit
$$\lim_{x\to\infty} x^2 [1-f(x)]$$
where
$$f(x)=\begin{cases}1-\frac{1}{2}\left(1-\frac{x}{\delta} \right)^p & \text{if } 0 \leq x \leq \delta \\ 1 & \text{if } x\geq \delta \end{cases}$$
I don't know how to carry out rigorously the exercise. In my attemp I have tried to study separately di $x^2$ and $1-f(x)$ for $x\to\delta$. Since $1-f(x)\to 0$ like $1-(1-x^p/\delta)\to 0$, my conclusion is that for $p\geq 2$ the limit is $0$, but I don't know if is correct.
On
Your result is correct, but your reasoning is not.
If you have that $x^2 \to \infty$ and $1 - f(x) \to 0$, you cannot conclude that $\lim_{x\to\infty} x^2(1- f(x)) = 0$. Consider the counterexample $x^2 \cdot \frac{1}{x} \to \infty$.
However, $1 - f(x) = 0$ for $x$ large enough, thus you can write
$$\lim_{x\to\infty} x^2(1 - f(x)) = \lim_{x\to\infty} x^2 0 = 0.$$
In the same vein:$$\lim_{x\to0} \frac{0}{x} = 0$$
Hint:
For large enough $x$ we have that $x \ge \delta$ so $1 - f(x) = 0$. Therefore $$\lim_{x\to\infty} x^2[1-f(x)] = \lim_{x\to\infty} 0 = 0$$