Squeeze theorem applied to x^2 + x sin(1/x)

Since $-1 \le \sin(1/x) \le 1$ for all $x \neq 0$, multiplying by $|x|$ gives $-|x| \le x\sin(1/x) \le |x|$, which holds for both positive and negative $x$. Adding $x^2$ to every part of this inequality yields $x^2 - |x| \le g(x) \le x^2 + |x|$, giving valid squeezing bounds near $x=0$.

Now take the limit of the outer expressions as $x \to 0$: both $x^2 - |x|$ and $x^2 + |x|$ approach $0 - 0 = 0$ by direct substitution, since these are continuous functions. Because $g(x)$ is squeezed between two functions that both tend to $0$, the Squeeze Theorem guarantees $\lim_{x\to 0} g(x) = 0$.

Both bounding functions, x squared minus the absolute value of x and x squared plus the absolute value of x, approach zero as x approaches zero. This shared value is exactly what forces the squeezed function g of x to also approach zero. The answer is zero.