Limit as $x \to 0$ of $\big(x \int_x^1{\frac{f(t)}{\sin^2(t)}}\: dt\big)$ Suppose $f$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$. Find:
$$\lim_{x\rightarrow{0^{+}}}{\bigg(x\int_x^1{\frac{f(t)}{\sin^2(t)}}\: dt\bigg)}$$
I am trying to use l'hopital's rule with:
$$\lim_{x\rightarrow{0^{+}}}{\bigg(\frac{\int_x^1{\frac{f(t)}{\sin^2(t)}}\: dt}{\frac{1}{x}}\bigg)}$$
However this only works if the integral evaluates to $\pm$infinity. when $x=0$. Since $f(t)$ is general I'm not really sure what to do next. 
Help would be appreciated :)
 A: Hint: L'Hopital is applicable in the case where the denominator $\to \infty,$ no matter what the numerator is doing.
A: Define $m(x)=\inf_{t\in[0,x]}t^2\frac{f(t)}{\sin^2(t)}$ and $M(x)=\sup_{t\in[0,x]}t^2\frac{f(t)}{\sin^2(t)}$. Since $f$ is continuous, then $m(x)\to f(0)$ and $M(x)\to f(0)$ as $x\to0^+$.
For $x,y\in(0,1)$, with $y<x$, there is $c\in(x,y)$ such that $$\frac{\int_{y}^{x}\frac{f(t)}{\sin^2(t)}}{\frac{1}{x}-\frac{1}{y}}=\frac{\int_{x}^{1}\frac{f(t)}{\sin^2(t)}-\int_{y}^{1}\frac{f(t)}{\sin^2(t)}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{f(c)}{\sin^2(c)}}{\frac{1}{c^2}}$$
Therefore, $$m(x)\leq\frac{y\int_{x}^{1}\frac{f(x)}{\sin^2(x)}-y\int_{y}^{1}\frac{f(y)}{\sin^2(y)}}{\frac{y}{x}-1}\leq M(x)$$
Taking $\liminf$ and $\limsup$ as $y\to0^+$ we get $$m(x)\leq\liminf_{y\to0^+}(\text{ or }\limsup_{y\to0^+}) y\int_{y}^{1}\frac{f(y)}{\sin^2(y)}\leq M(x)$$
Therefore, taking $x\to0^+$ we get that $\liminf(\text{ or }\limsup) y\int_{y}^{1}\frac{f(y)}{\sin^2(y)}=f(0)$.
A: If one assumes that $f$ is constant, then the integral evaluates to $f(0)(\cot x-\cot 1)$ and thus the desired limit is $f(0)$.
This motivates us to prove that the limit is $f(0)$ even when $f$ is not constant. We need only continuity of $f$ at $0$. Let $\epsilon>0$ then there is a $\delta >0$ such that $$|f(x) - f(0)|<\epsilon$$ whenever $|x|<\delta$.
We deal with the case when $x\to 0^+$ (the case $x\to 0^-$ being similar). We have $$x\int_{x}^{1}\frac{f(t)} {\sin^2t}\, dt=x\int_{x} ^{\delta} \frac{f(t)} {\sin^2t}\,dt+x\int_{\delta} ^{1}\frac{f(t)}{\sin^2t}\,dt$$ As noted in the start of the answer we have $$x\int_{x} ^{\delta} \frac{f(0)}{\sin^2t}\,dt=f(0)(x\cot x-x\cot\delta) $$ and hence $$\left|x\int_{x} ^{1}\frac{f(t)}{\sin^2t}\,dt-f(0)(x\cot x - x\cot\delta)\right|\leq x\int_{x} ^{\delta}\frac{|f(t)-f(0)|}{\sin^2t}\,dt+x\int_{\delta}^{1}\frac{|f(t)|}{\sin^2t}\,dt$$ and the RHS does not exceed $$\epsilon (x\cot x-x\cot \delta) +x\int_{\delta} ^{1}\frac{|f(t)|}{\sin^2t}\,dt$$ Taking limits as $x\to 0^+$ we see that $$f(0)-\epsilon\leq \liminf_{x\to 0^{+}} x\int_{x} ^{1}\frac{f(t)}{\sin^2t}\,dt\leq\limsup_{x\to 0^{+}} x\int_{x} ^{1}\frac{f(t)}{\sin^2t}\,dt\leq f(0)+ \epsilon $$ Since $\epsilon $ is an arbitrary positive number the desired limit is $f(0)$.
