Math / Calculus

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

STEP 1

1. The function $f(x) = x^2 \cos \frac{1}{x}$ is defined for $x \neq 0$ .
2. We need to demonstrate the inequality $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$ .
3. We will use the Sandwich Theorem to find the limit as $x \to 0$ .

STEP 2

1. Sketch the graph of $y = f(x)$ using a graphing calculator.
2. Prove the inequality $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$ .
3. Apply the Sandwich Theorem to find the limit as $x \to 0$ .

STEP 3

Use a graphing calculator to sketch the graph of $y = f(x) = x^2 \cos \frac{1}{x}$ .
Observe that as $x$ approaches 0, the oscillations of $\cos \frac{1}{x}$ become more frequent, but the amplitude is controlled by $x^2$ .

STEP 4

We know that for any real number $\theta$ , $-1 \leq \cos(\theta) \leq 1$ .
Thus, multiplying throughout by $x^2$ (which is non-negative), we get: $-x^2 \leq x^2 \cos \frac{1}{x} \leq x^2$

STEP 5

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:
$\boxed{0}$

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