Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x  \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x  \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\  \left({x \over x^2+x}+{x  \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x  \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$ 
Can we do this without squeeze theorem?

With a Riemannian sum:
$$\sum_{k=1}^n\frac{n}{n^2+k}=n\frac 1{n^2}\sum_{k=1}^n\frac1{1+\dfrac k{n^2}}$$ can be seen as a Riemaniann sum truncated to the $n$ first terms among $n^2$. Then
$$\lim_{n\to\infty}n\int_0^{1/n}\frac{dx}{1+x}=\lim_{n\to\infty}n\log\left(1+\frac1n\right)=1.$$
No quite rigorous, though (because the sum converges to the integral at the same time that we increase $n$).
