Determine the limit $$\lim_{x\to 0} x\int_x^1\frac{\cos t}{t^2}\mathrm{d}t $$
I would use the l'Hospital's rule, but for that I need to be able to justify that the integral tends to $\infty$. Not concerned with the calculation itself, rather need to justify the use of l'Hospital's rule.
Calculating out the series looks to be very tedious, no doubt it will answer the question, but is there a faster way?
On
Hint. You may observe that, as $x \to0$, $$ x\int_x^1\frac{\cos t}{t^2}\mathrm{d}t=x (1-x)\times \frac{\int_x^1\frac{\cos t}{t^2}\mathrm{d}t}{1-x}\sim x (1-x)\times \left(-\frac{\cos x}{x^2}\right) $$ and you may conclude easily.
We have used
$$ \frac{f(x)-f(a)}{x-a}\to f'(a) $$
with $f'(x)=-\dfrac{\cos x}{x^2}$ and $a=1$.
On
You may avoid De l'Hopital by noticing that $\frac{\cos t}{t^2} = \frac{1}{t^2}-\frac{2\sin^2(t/2)}{t^2}$, where the last part is a continuous and bounded function on $(0,1]$, hence, by exploiting $\int\frac{dt}{t^2}=-\frac{1}{t}$: $$ \int_{x}^{1}\frac{\cos t}{t^2}\,dt = -1+\frac{1}{x}-C+O(x) $$ (where $C=\int_{0}^{1}\frac{2\sin^2(t/2)}{t^2}\,dt$) and that clearly implies: $$ \lim_{x\to 0}x\cdot\int_{x}^{1}\frac{\cos t}{t^2}\,dt = \color{red}{1}.$$
$$\cos(t)=1+o(1)\implies \frac{\cos(t)}{t^2}=\frac{1}{t^2}+o\left(\frac{1}{t^2}\right)\implies \frac{\cos(t)}{t^2}\sim_0\frac{1}{t^2} \implies \int_x^1\frac{\cos(t)}{t^2}\mathrm d t=+\infty.$$ Therefore $$\lim_{x\to 0}\int_x^1 \frac{\cos(t)}{t^2}\mathrm d t=\lim_{x\to 0}\frac{\int_x^1 \frac{\cos(t)}{t^2}\mathrm d t}{\frac{1}{x}}\underset{Hospital}{=}...=1.$$
The fact that you can use l'Hospital is because you have an indetermination of the type $\frac{\infty }{\infty }$.