Find $\lim_{x \to 0^+}x\int_x^1 \frac{\cos t}{t^2}\,dt$ Find
$$\lim_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\,dt$$
This looks like an interesting problem,but i cannot figure out where to start, can anyone explain
 A: $$\cos(t)=1+O(t^2)$$
$$\frac{\cos(t)}{t^2}=\frac{1}{t^2}+O(1)$$
$$\int_{x}^1\frac{\cos(t)}{t^2}dt=-1+\frac{1}{x}+O(1)$$
$$x\int_{x}^1\frac{\cos(t)}{t^2}dt=-x+1+O(x)$$
$$\lim\limits_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt=1$$
A: You may start like this :
$$\lim\limits_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt =  \lim\limits_{x \to 0^+}\dfrac{\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt}{1/x}$$
and see if it satisfies the hypothesis for applying L'Hopital's rule
A: Substituting $s = \frac{1}{t}$ and $y = \frac{1}{x}$ you get
$$\lim_{x \rightarrow 0^+} x \int_x^1 \frac{\cos t}{t^2}dt = \lim_{x \rightarrow 0^+}
x\int_1^{\frac{1}{x}} \cos \left( \frac{1}{s} \right) ds = \lim_{y \rightarrow + \infty}
\frac{\int_1^y \cos \left( \frac{1}{s} \right) ds}{y}
$$
Now, applying De L'Hopital we get
$$\lim_{y \rightarrow + \infty} \frac{\int_1^y \cos \left( \frac{1}{s} \right) ds}{y}=
\lim_{y \rightarrow + \infty} \frac{\cos \left( \frac{1}{y} \right)}{1} = \cos0 = 1
$$
A: Substituting $u= \dfrac{t}{x}$, one has
$$x\int_x^1\frac{\cos t}{t^2}dt =  \int_{1}^{+\infty} \frac{\cos(ux)}{u^2} \chi_{\lbrace xu \leq 1\rbrace} du$$
Hence, Lebesgue's dominated convergence gives directly that
$$\lim_{x \rightarrow 0^+} x\int_x^1\frac{\cos t}{t^2}dt =  \int_{1}^{+\infty} \frac{1}{u^2} du$$
i.e. that $$\boxed{\lim_{x \rightarrow 0^+} x\int_x^1\frac{\cos t}{t^2}dt=1}$$
A: Integrating by parts we get $$\lim_{x\rightarrow 0+}x\int_{x}^{1}{\frac{\cos(t)}{t^{2}}dt}=\lim_{x\rightarrow 0+}\cos(x)-x\cos(1) -x\int_{x}^{1}{\frac{\sin(t)}{t}dt}$$ Since $\frac{\sin(x)}{x}$ extends continuously on $[0,1]$ we have that the Riemann-Integral $$\int_{0}^{1}{\frac{\sin(t)}{t}dt}$$ Exists, hence $$\lim_{x\rightarrow 0+}x\int_{x}^{1}{\frac{\sin(t)}{t}dt}=0$$
Thus $$\lim_{x\rightarrow 0+}x\int_{x}^{1}{\frac{\cos(t)}{t^{2}}dt} =1$$
