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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.

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  • $\begingroup$ Are you using $B$ to represent $\delta$ = \delta? $\endgroup$
    – amWhy
    Nov 8 '13 at 14:58
  • $\begingroup$ Yes, the definition is: there is a B so x<B => |f(x)-L|<E $\endgroup$
    – Michael
    Nov 8 '13 at 15:00
  • $\begingroup$ Your definition is logically wrong. "For any $\epsilon>0$" always comes at the beginning, never at the end. Moreover, it is not legitimate to assume $\delta=0$, since you want $\delta>0$. However, you can assume that $\delta$ is "small enough", and finally select the minimum value of $\delta$. $\endgroup$
    – Siminore
    Nov 8 '13 at 15:07
  • $\begingroup$ $\epsilon$ = \epsilon, and $\delta$ = \delta $\endgroup$
    – amWhy
    Nov 8 '13 at 15:07
  • $\begingroup$ @Siminore I edited the post, epsilon is at the begining now. But $\delta$ does not have to be > 0. But you can assume it. $\endgroup$
    – Michael
    Nov 8 '13 at 15:12
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What we want in the definition is

For any $\epsilon > 0$, there is a $\delta$ such that $x \lt \delta \implies |f(x) - L| \lt \epsilon.$

(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 $\delta$.

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  • $\begingroup$ $\delta$ doesn't have to be >0 but you can assume it. There are people who add it to the definition and others don't. Thanks for answering, could you please answer my second question? $\endgroup$
    – Michael
    Nov 8 '13 at 15:14
  • $\begingroup$ It depends what you mean when you say "assume many things." Often times, we can design each $\delta$ to be a function of $\epsilon$ so that for any $\epsilon >0$, we can ensure the existence of a $\delta$ that makes the implication true. $\endgroup$
    – amWhy
    Nov 8 '13 at 15:21

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