# How do you use the epsilon delta definition to find the limit of (2n) / (3n+1) as x approaches oo?

Nov 13, 2016

You don't.

#### Explanation:

Find ${\lim}_{x \rightarrow \infty} \frac{2 x}{3 x - 1}$

${\lim}_{x \rightarrow \infty} \frac{2 x}{3 x - 1} = {\lim}_{x \rightarrow \infty} \frac{2}{3 - \frac{1}{x}} = \frac{2}{3 - 0} = \frac{2}{3}$

There is no epsilon-delta definition for limits at infinity. A common definition is:

${\lim}_{x \rightarrow \infty} f \left(x\right) = L$ $\text{ }$ if and only if

for every positive $\epsilon$, there is an $M$ such that for every $x > M$, we have $\left\mid f \left(x\right) - L \right\mid < \epsilon$.

To prove that the limit is $\frac{2}{3}$, we need to show that the definition is satisfied.

In this case, let $M = \frac{2}{3 \epsilon}$.

For every $x > \frac{2}{3 \epsilon}$, we have $3 x + 1 > x > \frac{2}{3 \epsilon}$.

Therefore, for every $x > \frac{2}{3 \epsilon}$, we have $\epsilon > \frac{2}{3 \left(3 x + 1\right)}$ and,

since abs ((2x)/(3x-1)-2/3) = abs ((-2)/(3(3x+1)), we have

$\left\mid \frac{2 x}{3 x - 1} - \frac{2}{3} \right\mid < \epsilon$