I need to prove the following limit using definition only:
$$\lim_{x\to -\infty} \frac{(7x+3)}{(x-1)} =7.$$
The definition is: for any $\epsilon >0$. there is a $\delta$ so $x<\delta \implies |f(x)-L|<\epsilon$
In order to show that $|f(x) – L| < \epsilon$ I assumed $\delta =1, \delta=0$ and in the end i showed $\delta= 10/\epsilon$. The problem is I don't know which one to choose: $\min\{\delta_1, \delta_2, \cdots\}$ or $\max\{\delta_1, \delta_2, \cdots\}$. In the normal definition of a limit i know that we always choose the minimum delta but i am not sure what do here.
My second question is: can I assume many things about $\delta$ then just choose minimum/maximum, is it ok?
Thanks for help.
Best Answer
What we want in the definition is
(Definition since corrected in the OP).
For any $\epsilon > 0$, we can make $\delta$ as small as we'd like to ensure the implication holds, and in the end, we select the minimum $\delta$.