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How to show that $$ \int_0^1 \left\vert 1- \frac{\mathrm e^{-h(1+s^2)}-1}{-h(1+s^2)} \right\vert \mathbf ds \xrightarrow[h\to 0]{} 0$$ without using dominated convergence theorem? Does exists an "epsilon" proof or other elementary proof?

Thanks for any answers...

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    $ h \longmapsto 1- \frac{\mathrm e^{-h(1+s^2)}-1}{-h(1+s^2)} $ is continuous on $[0,1]$ as $\lim_{x \to 0} \frac{e^x- 1}{x} = 1$. Therefore consider $x_n \longrightarrow 0$ and use the uniform convergence of $$ f_n \longmapsto 1- \frac{\mathrm e^{-x_n(1+s^2)}-1}{-x_n(1+s^2)}$$. – 2017-01-09

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Hint: Observe \begin{align} 1-\frac{1-e^{-h(1+s^2)}}{h(1+s^2)}= \frac{h(1+s^2)-1+e^{-h(1+s^2)}}{h(1+s^2)}. \end{align} Using Lagrange remainder theorem(or Taylor Expansion), we see that \begin{align} e^{-x} =1-x+ e^{-\xi(x)}\frac{x^2}{2!} <1-x+\frac{x^2}{2}. \end{align} where $0<\xi(x)