Show that $\lim_{x \to 1} [2x + 1] \ne 10$ using the epsilon delta definition of a limit not existing

Question

I started with:

=> $|2x + 1 - 10| \ge \epsilon$

=> $|2x - 9| \ge \epsilon $

=> $ |2(x - 1) - 7| \ge \epsilon $

$ 0 < |x - 1| < \delta $ must be satisfied.

I'm not sure how to proceed after this. In the epsilon-delta proof of a limit we set $\delta$ to a value which satisfies $|f(x) - L| < \epsilon $ for all values of $\epsilon$. However I'm not sure if I should be setting a value of $\epsilon$ for the negation of a limit rather than a $\delta$, and if I do I'm not sure how to proceed with it to show that every $\delta$ value satisfies the $\epsilon$ value that I would have selected.

Answer 1

You want to show a limit does not exist. Therefore, you should show that there is some $\epsilon > 0$ for which you cannot find a $\delta>0$ so that the limit definition works out.

Now, what can this $\epsilon$ be? From what you have written above, we want that $|2(x-1) - 7| > \epsilon$. But then, we know that $x \to 1$, and so the LHS actually goes to seven, not zero. Therefore, any $0 < \epsilon < 7$ will do the job. Let's take $\epsilon = 6$.

We want to show that no $\delta > 0$ is such that if $0<|x-1| < \delta$ then $|2(x-1)-7| < 6$. That is, we have to find an $x$ such that $0<|x-1| < \delta$ but $|2(x-1)-7| > 6$. However, take $x = 1 - \frac \delta 2$, then $0<|1-x| < \delta$ and $|2(x-1)-7| = |-\delta - 7| = \delta+7 > 6$, clearly, since $\delta > 0$. Since the above holds for any $\delta > 0$, no $\delta$ can satisfy the limit definition for $\epsilon = 6$. Hence, the limit is not true.

Answer 2

For every $\delta > 0$, if $x := \frac{\delta}{2} + \frac{10}{2}$, then $|2x-9| = \delta+1 > 1 =: \varepsilon$. This proves the existence of the desired kind of $\varepsilon > 0$. For the intuition, just draw a graph to see if you can write the intuition out formally as the first sentence.