Integral - what is it ? (a little theoretical question). I have a little question, that run threw my thoughts, when i saw this exercise: $$\lim _{x\to \infty }\left(\frac{\int _{sin\left(x\right)}^xe^{\sqrt{t^2+1}}dt}{e^{\sqrt{x^2-1}}}\:\right)$$
Of course I want to implement here Lophital's rule, but without showing a calculation, is there intuitive and logical explanation why is the nominator (the integral) is $\infty$ ?
I asked myself this question, because I remember that double integral on constant 1 for exmaple is the the area, and triple on 1 is the velocity, so here I tried to think what is an integral with the same domain but with constant 1? 
$$\int _{sin\left(x\right)}^x1dt\:$$
That is defintly diverges because $x-sin(x)$ when x "aproaches" infnity.
And here, as I conclude, there isn't too much geometric meaning to that kind of integral, where the domain is between functions (like here x and sin x), I mean I can't say that it is the area that bounded by x and sin(x), the integral on 1 is just the result of substraction, right ? 
So, I dont have any geometrical intuition on that integral, except one thing I see clearly: $$$$ As I said, the integral on 1 diverges and the function $e^{\sqrt{x^2+1}}$ is defintly increasing function, but it is enough to say that it is bigger then 1 for each $x$ and that's why (one of the reasons the limit is $\infty$.

You can see that
$$\int_{\sin x}^x e^{\sqrt{t^2-1}} dt = \int_0^x e^{\sqrt{t^2-1}} dt + \int_{\sin x}^0 e^{\sqrt{t^2-1}} dt$$
and clearly the first summand diverges to $\infty$, while the second summand is bounded becuase
$$\left| \int_{\sin x}^0 e^{\sqrt{t^2-1}} dt \right| \le  \int_{-1}^1 e^{\sqrt{t^2-1}} dt = constant$$
So, in order to compute your limit, you can simply replace $\sin x$ with $0$, and then apply L'Hopital.
