If we know Taylor's series of $\mathrm{csch}^{2}(x)$ as below
$$\mathrm{csch}^2x= {\frac{1}{x^2}-\frac{1}{3}+\frac{x^2}{15}-\frac{2x^4}{189}+\frac{x^6}{675}+O(x^8)}$$
We want to see a limit result by using Taylor series method.
$$\lim_{{u\to \infty}} \frac{\int_{0}^{u} x^{0.2}(\mathrm{csch}^{2}(x)-1/x^{2}) \space dx}{\int_{0}^{u} x^{0.8}(\mathrm{csch}^{2}(x)-1/x^{2}) \space dx } $$
Could we follow the way below by substituting $u=1/y$, which $y→0^+$ here?
Thus, we obtain the result $0$:
$$\lim_{{y\to 0^+}} \frac{\int_{0}^{1/y} x^{0.2}(-\frac{1}{3}+\frac{x^2}{15}-\frac{2x^4}{189}+\frac{x^6}{675}) \space dx}{\int_{0}^{1/y} x^{0.8}(-\frac{1}{3}+\frac{x^2}{15}-\frac{2x^4}{189}+\frac{x^6}{675}) \space dx }=0 $$
Thank you for kind comment.