Limit , will you use L'H rule ?

Wow! beautiful problem! We will use finding the area method to do this problem.

Observe that both the numerator and denominator tend to $\infty$ . Also, $f(x)$ is periodic with period $2\pi$ and $g(x)$ is periodic with period $1$ .

This means that we can partition the integrals into many parts of known area and add them up.

Let $x=2m\pi$ where $m$ is a positive integer.

$\Rightarrow \int _{ 0 }^{ x }{ f\left( t \right) dt } =m\int _{ 0 }^{ 2\pi }{ f\left( t \right) dt } =m{ \pi }^{ 2 }$ (easily obtained from graph)

Now,

$\Rightarrow \int _{ 0 }^{ x }{ g\left( t \right) dt } =x\int _{ 0 }^{ 1 }{ g\left( t \right) dt } =x\int _{ 0 }^{ 1 }{ \sqrt { t } \sqrt { 1-t } dt } \\ =2x\int _{ 0 }^{ 1/2 }{ \sqrt { t } \sqrt { 1-t } dt } \\ =x\int _{ 0 }^{ 1/2 }{ \sqrt { 1-4{ t }^{ 2 } } dt } =x\frac { \pi }{ 8 }$

Thus:

${ \lim _{ x\rightarrow \infty }{ \frac { \int _{ 0 }^{ x }{ f\left( t \right) dt } }{ \int _{ 0 }^{ x }{ g\left( t \right) dt } } } }=\frac { 8m{ \pi }^{ 2 } }{ x\pi }$

$=\boxed{4}$ since $x=2m\pi$

Note: $\pi$ is an irrational number hence, $x=2m\pi$ is technically incorrect, but it doesn't matter in this case as we are making a small approximation with respect to huge terms.

Nice problem bro,

Well, for large, 'x' , we can neglect the fractional difference in area between the actual area and the area of the nearest complete loop

(example- if x = $\frac {3 \pi }{2}$ , then area is ${\pi }^{2} + \frac {{\pi }^{2}}{2}$

)

in general it is

$\left[ \frac { x }{ 2\pi } \right] { \pi }^{ 2 }+(some\quad fraction\quad of\quad { \pi }^{ 2 })\simeq \left[ \frac { x }{ 2\pi } \right] { \pi }^{ 2 }\simeq \frac { x }{ 2\pi } { \pi }^{ 2 }=\frac { x }{ 2 } { \pi }$ for large 'x'

similarly the other function is

$\left\lfloor x \right\rfloor \frac { \pi }{ 8 } +(some\quad fraction\quad of\quad \frac { \pi }{ 8 } )\simeq \frac { x }{ 8 } { \pi }$ for large 'x'

dividing both we get 4