Evaluating the difficult limit $\lim_{x\to 0^+} x \int^1_x \frac{f(t)}{\sin^2t}dt$ 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.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
