Proving limit to infinity using epsilon-delta definition.

Question

Given the following limit to infinity, I am supposed to prove using epsilon-delta definition that the limit = 3 as x approaches infinity. How do I approach this?

$$\lim_{x\to\infty}\dfrac{6x+1}{2x+1} = 3$$

Answer 1

$\big| \frac{6x+1}{2x+1}-3 \big| < \epsilon \iff \big| \frac{-2}{2x+1} \big| < \epsilon \iff \frac{1}{\epsilon}< |x+\frac{1}{2}|$.

Can you take it from here?

Answer 2

A limit to infinity for $f:\mathbb R\to\mathbb R$ (with value $l$) means that for all $\epsilon>0$ there is a $\delta>0$ such that if $|x|\geq\delta$ then $|f(x)-l|\leq\epsilon$.

Then you have:

Suppose $x>0$,

$$\big|\frac{6x+1}{2x+1}-3\big|=\frac{2}{|2x+1}|\leq\frac{2}{2|x|}\leq\frac{1}{\delta}.$$

Then you choose $\delta=\frac{1}{\epsilon}$ and you are done.

Suppose instead $x<0$,

$$\big|\frac{6x+1}{2x+1}-3\big|=\frac{2}{|2x+1|}=\frac{2}{|1-2|x||}\leq\epsilon,$$

by choosing $\delta=\frac{1}{\epsilon}+\frac{1}{2}$.