Can I break this limit into individual terms?

No, you can't do this. Here's a simpler example:

$$\lim_{n\to\infty} \frac1n + \frac1n + \cdots + \frac1n$$

where there are $n$ terms in the sum. If we add up the terms, then of course we have $\lim_{n\to\infty} 1 = 1$, even though each term is going to $0$.

The key is that the number of terms grows as $n$ tends to infinity. If there were a fixed number of terms, then you could take the sum of the limits of the terms.

Even if we speak about sequences, it is not possible to treat each term separately. The very simple example is the sequence $$\left(1+\frac{1}{n}\right)^n=\underbrace{\left(1+\frac{1}{n}\right)\cdot\left(1+\frac{1}{n}\right)\cdots\left(1+\frac{1}{n}\right)}_{n\text{ factors}}.$$ Each factor tends (separately) to $1$. Is it true that the limit is also $1$? No, this is $\text{e}$.

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

$$\lim_{x\to \infty}\left(\frac{x}{x^2+1}+\frac{x}{x^2+2}+\cdots +\frac{x}{x^2+x}\right)=1$$