Calculate a limit with an integral within proving that it's possible to use the L'Hôpital rule Let $f: [1, +\infty) \rightarrow R\;$ be a continuous function, bounded, and such that $f(x) \ge1 \;\;\;\forall\;x\ge1$. Calculate reasonably the following limit, proving that it is possible to use L'Hôpital Rule:
$$\lim_{x\to +\infty} \frac{1}{x} \int_{1}^{x^2} \frac{f(t)}{t}dt$$
I have been trying to prove we can use L'Hôpiatl rule by giving examples of functions which meet those conditions, such as the aditive polynomical, irrational (where the degree of the numerator is higher than the one of the denominator) and exponential functions, but then I'm stuck and I don't know how to continue.
Thank you! 
 A: Note that the denominator $x$ here tends to $\infty $ and thus L'Hospital's Rule can be applied. One should remember L'Hospital's Rule can be applied on two forms: "$0/0$" and "$\text{anything} /(\pm\infty) $".
Applying the rule here we see that limit in question is equal to the limit of $$\frac{f(x^2)}{x^2}\cdot 2x=2\cdot\frac{f(x^2)}{x}$$ provided the limit of above expression exists. Since $f$ is bounded the desired limit is $0$.

It is a common misconception that L'Hospital's Rule works on "$\infty/\infty $". One can see the emphasis on proving the limiting behavior of numerator in various answers here. This is entirely unnecessary.
If the denominator tends to $\infty$ or $-\infty $ then we can apply L'Hospital's Rule without worrying about limiting behavior of numerator. The rule will work if the expression obtained after differentiation of numerator and denominator tends to a limit.
A: For conditions under which L'Hospital is applicable refer to Paramanand Singh's answer.
L'Hospital:
Numerator: FTC and Chain Rule.
$\lim_{x \rightarrow \infty} \dfrac{\displaystyle{\int_{1}^{x^2}}(f(t)/t)dt}{x}=$
$\lim_{x \rightarrow \infty} \dfrac{(f(x^2)/x^2)(2x)}{1}=$
$\lim_{x \rightarrow \infty} 2\dfrac{f(x^2)}{x}.$
A: Substituting $t=u^2$ in the integral rewrites the limit as$$\lim_{x\to\infty}\frac{\int_1^x\frac{2f(u^2)du}{u}}{x}=\lim_{x\to\infty}\frac{2f(x^2)/x}{1}=2\lim_{x\to\infty}\frac{f(x^2)}{x},$$where we use the rule at the first $=$.
A: You can prove that
$$
\lim_{x\to\infty}\int_1^{x^2}\frac{f(t)}{t}\,dt=\infty
$$
because $f(t)/t\ge1/t$ and therefore
$$
\int_1^{x^2}\frac{f(t)}{t}\,dt\ge\int_1^{x^2}\frac{1}{t}\,dt=2\log x
$$
so the thesis follows by comparing limits.
Now you can apply l’Hôpital and the fundamental theorem of calculus (with the chain rule) to the form $\infty/\infty$ and get
$$
\lim_{x\to\infty}2x\frac{f(x^2)}{x^2}=\lim_{x\to\infty}2\frac{f(x^2)}{x}=0
$$
because $f$ is bounded.
