Q:

How to compute the $ \lim_{x \to 0+} x \int_x^1\frac{\cos t}{t^2}\, dt$?

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:

A simpler way is to apply the Hopital's rule to and then the Fundamental theorem of calculus:
$$\lim_{x→0+} \frac{\int_x^1\frac{\cos(t)}{t^2}\, dt}{1/x}=
\lim_{x→0+} \frac{-\frac{\cos(x)}{x^2}}{-1/x^2}=\lim_{x→0+}\cos(x)=1.$$

