Let $f$ be a function that is continuously differentiable twice on $[0, 1]$ such that $\int\limits_0^1 \frac{(f(x)-1)^2-4x^2}{x^\frac{7}{2}}dx$ converges.
Calculate $f(0), |f'(0)|$
*Note: we can use the fact that if $f$ is continuous on $[0, 1]$ and there exists $p \geq 1$ such that $\int\limits_0^1 \frac{f(x)}{x^p}dx$ converges
then $f(0)=0$
I used the note and found that $f(0)=1$, but I'm not sure how to find $|f'(0)|$
I thought about trying to use L'Hôpital together with the limit convergence test but it didn't work out.
I'm just looking for a hint on how to approach this.
Let $$ f(x)=1+f'(0)x+O(x^2)$$ and then \begin{eqnarray} &&\frac{(f(x)-1)^2-4x^2}{x^{\frac{7}{2}}}\\ &=&\frac{(f'(0)x+O(x^2))^2-4x^2}{x^{\frac{7}{2}}}\\ &=&\frac{(f'(0)+O(x))^2-4}{x^{\frac{3}{2}}}\\ &=&\frac{(f'(0))^2-4+O(x)}{x^{\frac{3}{2}}} \end{eqnarray} If $(f'(0))^2-4=0$ or $|f'(0)|=2$, then $$\frac{(f(x)-1)^2-4x^2}{x^{\frac{3}{2}}}=O(x^{-\frac12})$$ and hence $$ \int_0^1\frac{(f(x)-1)^2-4x^2}{x^{\frac{7}{2}}}dx $$ converges.