<s>[INST] 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?
 [/INST] Hint :
Take, $\displaystyle a_{x,n}=\frac{x^2}{x^2+n}$. Then use Cauchy's First limit theorem.
As $a_{x,n} \to 1$ when $x\to \infty$, so $\displaystyle \frac{1}{x}\sum_{n=1}^{x}a_{x,n} \to 1$ when $x\to \infty$.
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