Prove $\lim_{x \to 0} x^2\sin(1/x^2)/\sqrt x (x+1)=0$

Question

Prove $$\lim_{x \to 0} f(x)=0$$

Where $$f(x)=x^2\sin(1/x^2)/\sqrt x (x+1)$$

I have tried to prove $\lim\limits_{x \to 0}x^2\sin(1/x^2)=0$

Answer 1

Your limit doesn't exist as $f(x)$ is undefined for $x<0$. Although, right side limit does exist and is equal to $$\lim_{x\to0^+} \frac{x^2\sin(1/x^2)}{\sqrt x(x+1)}=\lim_{x\to0^+} \frac{x\sqrt x\sin(1/x^2)}{x+1}$$

The limit $\lim\limits_{x\to0^+}x\sqrt x\sin(1/x^2)$ is equal to $0$ because

$$-1\leqslant\sin(1/x^2)\leqslant1$$ $$-x\sqrt x\leqslant x\sqrt x\sin(1/x^2)\leqslant x\sqrt x$$

And $\lim\limits_{x\to0^+} x\sqrt x = 0$, so using the squeeze theorem we get $$\lim\limits_{x\to0^+}x\sqrt x\sin(1/x^2)=0$$

And finally,

$$\lim_{x\to0^+} \frac{x\sqrt x\sin(1/x^2)}{x+1}=0$$