On the following sequence (and similar ones): $\frac{2}{1},\frac{4}{3},\frac{6}{5},\dots \frac{2n}{2n+1}$, when we want to calculate the limit as $n\to\infty$, one usually divides both the numerator and the denominator by the highest power of the denominator.
Using the epsilon-delta definition, where this idea came from?
We want to be able to apply the Algebraic Limit Theorem. Specifically, we want to have two sequences $\{x_n\}$ and $\{y_n\}$ such that \begin{align} \lim_{n \to \infty} x_n &= x \\ \lim_{n \to \infty} y_n &= y, \end{align} where $\{y_n\}$ is eventually nonzero and $y \ne 0$, so that we can conclude \begin{align} \lim_{n \to \infty} \frac{x_n}{y_n} &= \frac{x}{y}. \end{align} The problem is that the sequence \begin{align} \frac{2n}{2n+1} \end{align} does not immediately satisfy this condition, since \begin{align} \lim_{n \to \infty} 2n &= \infty \\ \lim_{n \to \infty} 2n + 1 &= \infty. \end{align} However, by dividing both the numerator and the denominator by $2n$ ($n > 0$), we obtain \begin{align} \frac{2n}{2n+1} &= \frac{1}{1+ \frac{1}{2n}}. \end{align} Now, we have \begin{align} \lim_{n \to \infty} 1 &= 1, \\ \lim_{n \to \infty} 1 + \frac{1}{2n} &= \lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{1}{2n} \\ &= 1 + 0 \\ &= 1. \end{align} Here we use the Algebraic Limit Theorem to establish that $\lim_{n \to \infty} 1 + \frac{1}{2n} = 1$, but in a much more straightforward way. Notice, however, that these two results allow us to conclude that \begin{align} \lim_{n \to \infty} \frac{2n}{2n + 1} &= \lim_{n \to \infty} \frac{1}{1+ \frac{1}{2n}} \\ &= \frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty} 1 + \frac{1}{2n}} \\ &= \frac{1}{1} \\ &= 1, \end{align} as we needed.