Question1. Let $$f(x)=x^{2} \cos \frac{1}{x}, \quad x \neq 0$$ (a) Use a graphing calculator to sketch the graph of $y=f(x)$. (b) Show that $$-x^{2} \leq x^{2} \cos \frac{1}{x} \leq x^{2}$$ holds for $x \neq 0$. (c) Use your result in (b) and the sandwich theorem to show that $$\lim _{x \rightarrow 0} x^{2} \cos \frac{1}{x}=0$$

Studdy Solution
By the result in Step 2, we have: $$-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$$
As $x \to 0$, both $-x^2$ and $x^2$ approach 0.
By the Sandwich Theorem, since $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$ and both bounds approach 0, we conclude: $$\lim_{x \to 0} x^2 \cos \frac{1}{x} = 0$$
The limit is:
$$\boxed{0}$$