<s>[INST] Can I break this limit into individual terms? $$\lim_{x\to \infty}  {\frac{x}{x^2+1} +\frac{x}{x^2+2} + ... + \frac{x}{x^2+x} }$$
It seems obvious that the result is zero for each term but in order to break the limit into its individual  parts we must know that every term's limit exists . 
 [/INST] Note that the series of interest can be written as
$$\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}=\sum_{k=1}^x \frac{x}{x^2+k}$$
The summand clearly satisfies the bounds 
$$\frac{x}{x^2+x}\le\frac{x}{x^2+k}\le \frac{x}{x^2+1}$$
Hence we can assert that 
$$\frac{x^2}{x^2+x}\le \sum_{k=1}^x\frac{x}{x^2+k}\le \frac{x^2}{x^2+1}$$
whereby application of the squeeze theorem yields the coveted limit
$$\lim_{x\to \infty}\left(\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}\right)=1$$
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