Convergence of $f_n(x)=n^2f(nx)$ in the sense of distributionas

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Let $f$ be a test function such that $\int_{-\infty}^\infty f(x)dx=0$ and $f_n(x)=n^2f(nx)$. Find the distributional limit $\lim_{n\to\infty}f_n$.

How can I use the Dominated Convergence Theorem here? I have problems using the conditions that the integral is zero. Can someone give me a hint?

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Let $g$ be smooth and compactly supported. Write $u=nx$ so that

$$\int_{-\infty}^\infty n^2 f(nx) g(x) dx = \int_{-\infty}^\infty n f(u) g(u/n) du.$$

Now by the hypotheses, $g_n(u) := g(u/n)$ converges uniformly to $g(0)$. (We needed the compact support for this.) This motivates splitting the integral by adding and subtracting $g(0)$ as follows:

$$\int_{-\infty}^\infty n f(u) g(u/n) du = \int_{-\infty}^\infty n f(u) (g(u/n)-g(0)) du + \int_{-\infty}^\infty n f(u) g(0) du.$$

The second term is zero regardless of $n$ by the zero integral hypothesis. Now to deal with the first term, write the multiplication by $n$ as a division by $1/n$:

$$\int_{-\infty}^\infty n f(u) (g(u/n) - g(0)) du = \int_{-\infty}^\infty u f(u) \frac{g(u/n) - g(0)}{u/n}.$$

Then the ratio converges uniformly to $g'(0)$. What happens now?