I am working out the proof that
$$ \lim_{→0+} x \int_x^1\frac{\cos t}{t^2}\, dt =1 $$
My attempt is firstly handle the integral parts:
$$ \int_x^1\frac{\cos t}{t^2}\, dt = \left(-\left(-\text{Si}(x)-\frac{\cos (x)}{x}\right)-\text{Si}(1)-\frac{\cos (1)}{1}\right. $$
where $$ \text{Si}(z)=\int _0^z\frac{\sin(t)}{t}\, dt $$
Then $$x \int_x^1\frac{\cos t}{t^2} = xSi(x) + \cos(x)-xSi(1)-x\cos(1)$$ so, $$\lim_{→0+} x \int_x^1\frac{\cos t}{t^2} = \lim_{→0+} (xSi(x) + \cos(x)-xSi(1)-x\cos(1))$$
And also, $$ \lim_{→0+} xSi(1) = 0$$ $$\lim_{→0+}x\cos(1)=0 $$ $$\lim_{→0+}\cos(x)=1 $$ $$\lim_{→0+} xSi(x)=0 ??$$ $$$$
then $$\lim_{→0+} x \int_x^1\frac{\cos t}{t^2} = 1$$
I am uncertain about my proof. Is it correct? Thanks in advance.
A simpler way is to apply the Hopital's rule to and then the Fundamental theorem of calculus: $$\lim_{x→0+} \frac{\int_x^1\frac{\cos(t)}{t^2}\, dt}{1/x}= \lim_{x→0+} \frac{-\frac{\cos(x)}{x^2}}{-1/x^2}=\lim_{x→0+}\cos(x)=1.$$