Check pointwise convergence and uniform convergence of $f_n(x) = n^3x\exp(-nx^2)$ on $[0,1]$
Pointwise convergence: $$ \lim_{n\rightarrow\infty} f_n(x) = \lim_{n\rightarrow\infty} n^3x\exp(-nx^2) =\lim_{n\rightarrow\infty} \exp(ln(n^3x))\exp(-nx^2) = \\ = \lim_{n\rightarrow\infty} \exp(\ln(n^3x)-nx^2) = \lim_{n\rightarrow\infty} \exp(3\ln(n) + ln(x) - nx^2) = 0$$
so if $f_n$ converges uniformly to $f$ then for any $\varepsilon > 0$ there is $n_0$ that for any $n>n_0$ and $x \in \mathbb R$ $$ \exp(3\ln(n) + ln(x) - nx^2) < \varepsilon $$ so I choose $\varepsilon = 1$ and I try to show that there is no $n_0$. Let $x = \frac1{\sqrt{n}}$ $$\exp(3\ln(n) + ln(\frac1{\sqrt{n}}) - 1) = \exp(3\ln(n) -\frac12\ln(n) - 1) = \exp(\ln(n^\frac52) - ln(e)) = \frac{n^\frac52}{e} > 1 $$ for $n>1$. So if I am right then $f_n$ convergence pointwise and do not convergence uniformly.
However, there is different answer in my book (convergence pointwise and uniformly), so where I am wrong?