Prove that if $\lim_{x\to\infty} \frac{f(x)}{g(x)}=1$ then $\lim_{x\to\infty} \frac{\int_{x}^{\infty}f(t)dt}{\int_{x}^{\infty}g(t)dt}=1$ We are given:


*

*$f$ and $g$ are two positive and continuous functions on an interval $[a,\infty)$.

*$\int_{a}^{\infty}f(t)dt$ and $\int_{a}^{\infty}g(t)dt$ are convergent. 

 A: From $\lim_{x\to\infty} \frac{f(x)}{g(x)}=1$, we know for any $\epsilon > 0$. there exists $M > 0$, for all $x > M$, we have
$$ 1 -\epsilon < \frac{f(x)}{g(x)} < 1 + \epsilon $$
Or
$$(1 -\epsilon)g(x) < f(x) < (1 + \epsilon)) g(x)$$
Integrate it,
$$(1 -\epsilon)\int_x^{\infty}g(t)dt < \int_x^{\infty}f(t)dt < (1 + \epsilon))\int_x^{\infty}g(t)dt$$
Therefore,
$$(1 -\epsilon) < \frac{\int_x^{\infty}f(t)dt}{\int_x^{\infty}g(t)dt} < 1 + \epsilon$$
for x > M.
A: $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=1\implies\,\forall\,n\in\Bbb N\;\exists\,R_n\in\Bbb R^+\;\text{ such that for}\;\;x>R_n\;,\;\;\left|\frac{f(x)}{g(x)}-1\right|<\frac1n\iff$$
$$\iff g(x)\left(1-\frac1n\right)<f(x)<g(x)\left(1+\frac1n\right)\implies$$
$$1{}\xleftarrow[\infty\leftarrow n]{}1-\frac1n=\frac{\int\limits_x^\infty g(t)\left(1-
\frac1n\right)dt}{\int\limits_x^\infty g(t)dt}\le\frac{\int\limits_x^\infty f(t)dt}{\int\limits_x^\infty g(t)dt}\le\frac{\int\limits_x^\infty g(t)\left(1+\frac1n\right)dt}{\int\limits_x^\infty g(t)dt}=1+\frac1n\xrightarrow[n\to\infty]{}1$$
We can choose the above $\;R_n\;$ so as to have $\;x\to\infty\;$ when $\;n\to\infty\;$ . I think this poses no major problem.
A: Note that since the integrals $\int_a^\infty f(x)\,dx$ and $\int_a^\infty g(x)\,dx$ convege, then
$$\lim_{x\to \infty}\int_x^\infty f(t)\,dt=0$$
and
$$\lim_{x\to \infty}\int_x^\infty g(t)\,dt=0$$
Then, applying L'Hospital's rule, and applying the fundamental theorem of calculus, we find 
$$\lim_{x\to \infty}\frac{\int_x^\infty f(t)\,dt}{\int_x^\infty g(t)\,dt}=\lim_{x\to \infty}\frac{-f(x)}{-g(x)}=1$$
as was to be shown!
A: You can use L'Hopital but you can also prove it directly. Because of the limit given, for any $\epsilon > 0$ there's an $x_0$ such that if $t > x_0$ one has
$$(1 - \epsilon) g(t) < f(t) < (1 + \epsilon) g(t)$$
Integrating the above one obtains 
$$(1 - \epsilon)\int_{x_0}^{\infty} g(t)\,dt < \int_{x_0}^{\infty} f(t)\,dt < (1 - \epsilon)\int_{x_0}^{\infty} g(t)\,dt$$
This will hold if one replaces $x_0$ by any larger number too, so if $x > x_0$ one has
$$(1 - \epsilon)\int_{x}^{\infty} g(t)\,dt < \int_{x}^{\infty} f(t)\,dt < (1 - \epsilon)\int_{x}^{\infty} g(t)\,dt$$
This is the same as
$$(1 - \epsilon) < {\int_{x}^{\infty} f(t)\,dt \over \int_{x}^{\infty} g(t)\,dt} < (1 + \epsilon)$$
This gives the desired limit. 
