$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty} \frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}}\ dx$ Let 
$$f(x)=\frac{\tanh(\frac{\sqrt{x}}{n})}{x^2\sqrt{(n+1)x}} \ \ \ \text{and} \ \ f_n=f\chi_{[ 1/n^2,+\infty)}$$
I want to compute $$\lim_n \frac{1}{n}\int_{1/n^2}^{+\infty}f \ dx$$
My solution is the following:
Since $$\tanh(\frac{\sqrt{x}}{n})\leq \tanh(\sqrt{x}) \ \ \text{and} \ \ x^2\sqrt{(n+1)x} \geq x^2\sqrt{2x} \ \ \  \text{for} \ \ n\geq1$$
we have $$f_n\leq \frac{\tanh(\sqrt{x})}{x^2\sqrt{2x}} \ \ \text{for} \ \ x\in[1/n^2,1] \ \ n\geq 1$$
and $$f_n \leq \frac{1}{x^2} \ \ \text{for} \ x\in[1,+\infty)$$
Then by dominated convergence th. we have that the limit is equal to zero.
The solution given in my notes is much more involved than mine, so I guess I'm missing something.. 
Question

What is wrong with my solution?


Change variables with $$ y = \frac{\sqrt{x}}{n} \Rightarrow dy = \frac{dx}{2n\sqrt{x}}. $$
Also, $$ \sqrt{x} = ny \Rightarrow x^2 = n^4y^4. $$
When $ x = \frac{1}{n} $ we get that $ y = \sqrt{1 + \frac{1}{n^2}} \to 1.$
Finally, observe that $$ \int_1^{\infty} \frac{tanh(y)}{y^5} < + \infty$$ hence $$ \lim_{n \to \infty} \frac{1}{n} \int_{\frac{1}{n^2}}^{\infty} \frac{tanh\left( \frac{\sqrt{x}}{n}\right)}{x^2\sqrt{2x}} dx = 0. $$
