Find $\lim_{x \to 0^+}x\int_x^1 \frac{\cos t}{t^2}\,dt$
Solution 1:
$$\cos(t)=1+O(t^2)$$ $$\frac{\cos(t)}{t^2}=\frac{1}{t^2}+O(1)$$ $$\int_{x}^1\frac{\cos(t)}{t^2}dt=-1+\frac{1}{x}+O(1)$$ $$x\int_{x}^1\frac{\cos(t)}{t^2}dt=-x+1+O(x)$$ $$\lim\limits_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt=1$$
Solution 2:
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