# How do you prove the statement lim as x approaches 1 for [ x + 1 ] / [ 2x + 1 ] = 2/3 using the epsilon and delta definition?

Mar 24, 2016

To prove that ${\lim}_{x \to c} f \left(x\right) = L$ for some function $f$, we must show that for any $\epsilon > 0$ there exists a $\delta > 0$ such that if $| x - c | < \delta$ then $| f \left(x\right) - L | < \epsilon |$

Proof:
Let $\epsilon > 0$ be arbitrary, and let $\delta = \min \left(1 , 3 \epsilon\right)$. Then, if $| x - 1 | < \delta$, note that $| x - 1 | < 1$ and thus $| 2 x + 1 | > 1$ (see below for details). With that, we have, for $| x - 1 | < \delta$:

$| \frac{x + 1}{2 x + 1} - \frac{2}{3} | = | \frac{3 x + 3}{3 \left(2 x + 1\right)} - \frac{4 x + 2}{3 \left(2 x + 1\right)} |$

$= | \frac{- x + 1}{3 \left(2 x + 1\right)} |$

$= | \frac{x - 1}{3 \left(2 x + 1\right)} |$

$= | x - 1 \frac{|}{3 | 2 x + 1 |}$

$< | x - 1 \frac{|}{3}$

$< \frac{3 \epsilon}{3}$

$= \epsilon$

Therefore, as we have shown that a $\delta$ exists for any $\epsilon$ such that $| x - 1 | < \delta |$ implies $| \frac{x + 1}{2 x + 1} - \frac{2}{3} | < \epsilon$, it is the case that ${\lim}_{x \to 1} \frac{x + 1}{2 x + 1} = \frac{2}{3} \text{ }$

To see where we got $| 2 x + 1 | < 1$, starting from $| x - 1 | < 1$ we have:

$| x - 1 | < 1$

$\implies - 1 < x - 1 < 1$

$\implies - 2 < 2 x - 2 < 2$

$\implies 1 < 2 x + 1 < 5$

$\implies 1 < | 2 x + 1 | < 5$

$\therefore | 2 x + 1 | > 1$