Problem: Let $f:[0,\infty)\to\Bbb R$ be a continuous function. Assume that $$\lim_{x\to\infty}\frac{f(x)}{x^m}=1,\ \ m\in\Bbb N$$ Show that $$\lim_{x\to\infty}\frac{m+1}{x^{m+1}}\int\limits_0^xf(y)\, \mathrm{d}y=1.$$
Solution: Given $$\lim_{x\to\infty}\frac{f(x)}{x^m}=1$$ implies that for any $\varepsilon>0$, $\exists\alpha>0$ such that $\forall x>\alpha$ $$\left|\frac{f(x)}{x^m}-1\right|<\frac{\varepsilon}2\implies |f(x)-x^m|<\frac{\varepsilon}2x^m.$$ Now $$\left|\frac{m+1}{x^{m+1}}\int\limits_0^xf(y)\, \mathrm{d}y-1\right|=\left|\frac{m+1}{x^{m+1}}\left(\int\limits_0^\alpha f(y)\, \mathrm{d}y+\int\limits_\alpha^xf(y)\, \mathrm{d}y\right)-1\right|\leq\left|\frac{m+1}{x^{m+1}}\int\limits_0^\alpha f(y)\, \mathrm{d}y\right|+\left|\frac{m+1}{x^{m+1}}\int\limits_\alpha^xf(y)\, \mathrm{d}y-1\right|.$$ Since $f$ is continuous, $\exists M>0$ such that $|f(x)|\leq M$ for all $x\in[0,\alpha]$. So $$\left|\frac{m+1}{x^{m+1}}\int\limits_0^xf(y)\, \mathrm{d}y-1\right|\leq\frac{m+1}{x^{m+1}}\int\limits_0^\alpha|f(y)|\, \mathrm{d}y+\left|\frac{m+1}{x^{m+1}}\int\limits_\alpha^x(f(y)-y^m)\, \mathrm{d}y\right|+\frac{\alpha^{m+1}}{x^{m+1}}$$$$\leq\frac{M\alpha(m+1)+\alpha^{m+1}}{x^{m+1}}+\frac{m+1}{x^{m+1}}\int\limits_\alpha^x|f(y)-y^m|\, \mathrm{d}y<\frac{M\alpha(m+1)+\alpha^{m+1}}{x^{m+1}}+\frac{m+1}{x^{m+1}}\int\limits_\alpha^x\frac{\varepsilon}2y^m\, \mathrm{d}y<\frac{M\alpha(m+1)}{x^{m+1}}+1+\frac{\varepsilon}2(1-\frac{\alpha^{m+1}}{x^{m+1}})<\varepsilon$$ for all $x>\max\left\{\alpha,\left(\frac{2M\alpha(m+1)}{\varepsilon-2}\right)^{\frac1{m+1}}\right\}$.
Thus $$\lim_{x\to\infty}\frac{m+1}{x^{m+1}}\int\limits_0^xf(y)\, \mathrm{d}y=1.$$
Kindly check this solution and correct it if there is any flaw.
Thank you.
We are looking at
$$\tag 1 \frac{\int_0^x f(y)\,dy}{x^{m+1}/(m+1)}.$$
Now as $x\to \infty,$ the denominator $\to \infty.$ We are in the indeterminate form $\dfrac{?}{\infty}.$ Thus we can apply L'Hospital! For some strange reason a lot of students don't know this form of L'Hopital. It's just as easy to prove as the $\frac{\infty}{\infty}$ case.
Well, OK, suppose you don't know this and you're mired in the $\frac{\infty}{\infty}$ cult. You could proceed as Ty was trying to do. From the hypotheses, we know there exists $a>0$ such that $f(x)/x^m >1/2$ for $x>a.$ Thus for $x>a,$
$$\int_0^x f(y)\,dy > \int_a^x f(y)\,dy$$ $$>\int_a^x (1/2)y^m\,dy = \frac{x^{m+1}-a^{m+1}}{2(m+1)}.$$
The last term above $\to \infty$ as $x\to \infty.$ So the numerator in $(1)$ $\to \infty$ and "regular" L'Hopital applies.