Evaluating the difficult limit $\lim_{x\to 0^+} x \int^1_x \frac{f(t)}{\sin^2t}dt$

Question

I have for the past week been trying to determine the following limit theoretically with no success, supposing $f$ is differentiable, $$\lim_{x\to 0^+} x \int^1_x \frac{f(t)}{\sin^2t}dt$$ I have tried some polynomial values for $f$ and it appears to go to $f(0)$. However I have no idea as to why this is.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{x \to 0^{\large +}}\bracks{% x\int^{1}_{x}{\mrm{f}\pars{t} \over \sin^{2}\pars{t}}\,\dd t} & = -\lim_{x \to \infty}\bracks{% {1 \over x}\int_{1}^{1/x}{\mrm{f}\pars{t} \over \sin^{2}\pars{t}}\,\dd t} = -\lim_{x \to \infty}\bracks{% {\mrm{f}\pars{1/x} \over \sin^{2}\pars{1/x}}\,\pars{-\,{1 \over x^{2}}}} \\[5mm] & = \lim_{x \to \infty}\braces{% \bracks{1/x \over \sin\pars{1/x}}^{2}\,\mrm{f}\pars{1 \over x}} = \bbx{\lim_{x \to \infty}\mrm{f}\pars{1 \over x}} \end{align}