Question

a) Graph 𝑔(𝑥) = 𝑥 𝑠𝑖𝑛(1⁄𝑥) to estimate lim 𝑔(𝑥), zooming in on the origin as necessary 𝑥→0

(b) Confirm your estimate in part (a) with a proof.

Answer 1

(a)

$$\lim\limits_{x\to 0}g(x)=\lim\limits_{x\to 0}x\sin(1/x)=0$$

(b)

Use the Squeeze Theorem

$$-1\leq \sin(1/x)\leq 1, x\in\R, x\not=0$$

Then

$$-x\leq x\sin(1/x)\leq x, x\in\R, x>0$$

$$\lim\limits_{x\to 0^+}(-x)=\lim\limits_{x\to 0^+}x=0$$

By Squeeze Theorem

$$\lim\limits_{x\to 0^+}x\sin(1/x)=0$$

$$x\leq x\sin(1/x)\leq -x, x\in\R, x<0$$

$$\lim\limits_{x\to 0^-}(-x)=\lim\limits_{x\to 0^-}x=0$$

By Squeeze Theorem

$$\lim\limits_{x\to 0^-}x\sin(1/x)=0$$

We have

$$\lim\limits_{x\to 0^-}x\sin(1/x)=0=\lim\limits_{x\to 0^+}x\sin(1/x)$$

Therefore

$$\lim\limits_{x\to 0}x\sin(1/x)=0$$