Question : Let $f_n$ be the distribution $<f_n,\varphi>=n(\varphi(\frac{1}{n})-\varphi(\frac{-1}{n}))$. What distribution is $\lim_{n \to \infty} <f_n, \varphi>$ ?
First try : $\lim_{n \to \infty} n(\varphi(\frac{1}{n})-\varphi(\frac{-1}{n})) = \lim_{n \to \infty} \frac{\varphi(\frac{1}{n})-\varphi(\frac{-1}{n}))}{\frac{1}{n}}\stackrel{H.R}{=} \lim_{n \to \infty} \varphi'(\frac{1}{n})+\varphi'(\frac{-1}{n})$
Second try : $\lim_{n \to \infty} =n(\varphi(\frac{1}{n})-\varphi(\frac{-1}{n})) = \lim_{n \to \infty} n \int_{\frac{-1}{n}}^{\frac{1}{n}} \varphi'(x) dx$
How could I find $\lim_{n \to \infty} <f_n, \varphi>$?
Your approaches are both correct, you just have to finish them.
First try: this is the easiest method:
$$\lim \limits _{n \to \infty} \frac{\varphi \left( \frac 1 n \right) - \varphi \left( - \frac 1 n \right)} {\frac 1 n} = \lim \limits _{n \to \infty} \frac{\varphi \left( \frac 1 n \right) - \varphi (0)} {\frac 1 n} + \lim \limits _{n \to \infty} \frac{\varphi \left( - \frac 1 n \right) - \varphi (0)} {- \frac 1 n} = \\ 2 \varphi' (0) = \langle 2 \delta, \varphi' \rangle = \langle -2 \delta', \varphi \rangle .$$
Second try: this is more heavyweight since it involves the use of Lebesgue's dominated convergence theorem which is a conceptually subtle result:
$$\lim \limits _{n \to \infty} n \int \limits _{- \frac 1 n} ^{\frac 1 n} \varphi'(x) \ \Bbb dx = [\text{make } y = nx] = \lim \limits _{n \to \infty} \int \limits _{-1} ^1 \varphi' \left( \frac y n \right) \ \Bbb dy = \\ \int \limits _{-1} ^1 \lim \limits _{n \to \infty} \varphi' \left( \frac y n \right) \ \Bbb dy = \int \limits _{-1} ^1 \varphi' (0) \ \Bbb dy = 2 \varphi' (0) .$$
To conclude, $\lim \limits _{n \to \infty} f_n = -2 \delta'$.