# Find the limit and give the strict $\epsilon - \delta$ proof that: $\lim_{x\to 1^{+}}$ $x\lfloor x \rfloor$

This is from Advance Calculus Theory and Practice By John S. Petrovic. The book has the answer in the back but I need some clarity on $$\delta$$.

Answer: The limit is 1. Let $$\epsilon > 0$$, and choose $$\delta = \min{(1,\epsilon)}$$. If $$1 < x < 1 +\delta$$ then $$\lfloor x \rfloor = 1$$, so $$|x \lfloor x \rfloor - 1| = | x-1| < \epsilon$$. Where/how are they choosing the minimum? Does it have to do with $$\lfloor x \rfloor = 1$$ and $$| x-1| < \epsilon$$? Are they assuming a number and using the opposite to find the other?

• the minimum can be imposed because only small values of $\delta$ are important. For big values you can choose a random number and find an upper quote for $\epsilon$ Commented Jun 26 at 2:01
• They impose the min because then $\lfloor x \rfloor = 1$, so you get a simple formula for $x\lfloor x \rfloor-1 .$ Introducions to analysis are full of these simplifying "tricks". Commented Jun 26 at 2:12
• @hellofriends I wouldn't say it's because big $\delta$ is unimportant, it's because the prover has complete control over the value of $\delta$, and how small $\delta$ should be. The "big values are unimportant" reasoning applies to assumptions about how small $\varepsilon$ can be. Commented Jun 26 at 2:17
• @TheoBendit yes, that's what I wanted to say. I also meant lower quote for $\epsilon$ instead of upper (all the $\epsilon$ above this quote would satisfy the property) Commented Jun 26 at 2:20
• @hellofriends And I likewise meant to say "assumptions about how big $\varepsilon$ can be." :) Commented Jun 26 at 2:22

## 2 Answers

I speculate that the choice of $$~\delta \leq 1~$$ is designed to guarantee that $$~\left\lfloor x\right\rfloor = 1.~$$ As indicated by the comment of Steen, this guarantee simplifies the analysis.

So, with the constraint on $$~\delta~$$ that it is $$~\leq 1,~$$ the constraint that $$~1 < x < 1 + \delta~$$ does imply that $$~1 < x < 2,~$$ as intended.

• Thank you, I'm not sure I quite get it yet but this is helpful. I'm going to go try a few more!
– Cat
Commented Jun 26 at 14:08

Recall that once we have $$\epsilon$$ and $$\delta$$ we need it to be true that

$$\forall x : (0 < x - 1 < \delta \implies \lvert x \lfloor x \rfloor - 1\rvert < \epsilon). \tag1$$

(Since this is a one-sided limit, we don't take the absolute value of $$x - 1$$.)

Suppose we had the strategy to prove the existence of a suitable $$\delta$$ for every $$\epsilon$$ by simply setting $$\delta = \epsilon$$ without the $$\min$$ function.

Since the definition of a limit says "for all $$\epsilon > 0$$," a proof should cover all possible positive values of $$\epsilon$$. One such value of $$\epsilon$$ is $$2$$. Using the simple strategy $$\delta = \epsilon$$, when presented with $$\epsilon = 2$$ we would have $$\delta = 2$$. But then there exists a value of $$x$$, in particular $$x = 2.5$$, such that $$0 < x - 1 = 1.5 < \delta$$ but $$\lvert x \lfloor x \rfloor - 1\rvert = 4 \not< \epsilon$$. Therefore the implication in $$(1)$$ is not true for all $$x$$ and our proof strategy has failed.

Basically, if there is any way that $$\lfloor x \rfloor$$ can be greater than $$1$$, we're going to be in trouble on the right-hand side of the implication.

The strategy $$\delta = \min\{1,\epsilon\}$$ is a way to avoid this kind of failure. There are other ways the problem might be solved, but this is a very popular technique.