A function $f(x)$ is continuous at $x=0$ if the limit of the function as $x$ approaches $0$ is equal to the value of the function at $x=0$. That is: $$\lim_{x \to 0} f(x) = f(0)$$
We need to find $\lim_{x \to 0} x^{2}\sin(\frac{1}{x})$. Since $-1 \le \sin(\frac{1}{x}) \le 1$, we can use the Squeeze Theorem: $$-x^2 \le x^2\sin\left(\frac{1}{x}\right) \le x^2$$ As $x \to 0$, both $-x^2$ and $x^2$ approach $0$. Therefore, by the Squeeze Theorem, the limit is $0$.
From the definition of the function, $f(0) = k(0+1) = k$.
Equating the limit to the function value: $$0 = k$$ Thus, $k = 0$.
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