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?

Just using that $\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^2)$ and $\sum_{i=1}^n i=\frac{n(n+1)}{2}$
So $$\frac{x}{x^2+c}=\frac{1}{x}\frac{x^2}{x^2+c}=\frac{1}{x}\frac{1}{1+\frac{c}{x^2}}=\frac{1}{x}\left(1-\frac{c}{x^2}+O(\frac{c}{x^4})\right)$$
$$\sum_{c=1}^x\frac{x}{x^2+c}=\frac{1}{x}\sum_{c=1}^x1-\frac{c}{x^2}+O(\frac{c}{x^4})=1-\frac{x (x+1)}{2x^3}+O(\frac{1}{x^2})=1-\frac{1}{2x}+O(\frac{1}{x^2})$$
Note that using $\frac{1}{1+\epsilon}=1+O(\epsilon)$ is in fact enough and simpler here, but you don't get the $\frac{-1}{2x}$, only $O(\frac{1}{x})$
