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 .

Solution 1:

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.

Solution 2:

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 $\text{e}$.

Solution 3:

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

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