Limit / Integral

Hi, I was just wondering how would I go about calculating this limit. At first, I was thinking I can use L'Hopital's rule, but then I would be assuming that the integral tends to infinity as x tends to infinity and I'm not sure if I can just do that in my course, without calculating it. It is a methods course. We usually don't calculate integrals which look tough, but we prove whether they are convergent or divergent. Would it be enough to say that because this integral is divergent, it must have a value of +infinity as it has a non-negative integrand. I would show it is divergent by the Direct Comparison test with the function $\sqrt t^4$. Then, after when I use the L'Hopital's rule, it is clear I would use FTC on the top and the bottom is fine, etc..

Yeah all you really need to notice is that as $x \to \infty$, we have

$\displaystyle \int_0^x \sqrt{1+t^4} \ dt \sim \int_0^x \sqrt{t^4} \ dt \sim \dfrac{x^3}{3}$

No need for L'Hopitals.

Yeah all you really need to notice is that as $x \to \infty$, we have

$\displaystyle \int_0^x \sqrt{1+t^4} \ dt \sim \int_0^x \sqrt{t^4} \ dt \sim \dfrac{x^3}{3}$

No need for L'Hopitals.

Yeah I suppose this is also possible (though for rigour I would use greater than or equal to signs). It's still worth mentioning that there are limits of this type where it's necessary to use FTC.

Inequalities are not too important here. All we care about is the asymptotic behaviour as $x \to \infty$, so we just need to obtain a useful asymptotic expression for the integrand; doesn't really matter whether it's less than or greater than the expression we start with.