0

Guess the following limits and prove your answers using the epsilon-delta definition:

$ 1) \lim\limits_{x\to 2} (x^2-2x)$

$ 2) \lim\limits_{x\to2} \dfrac{5x+1}{2x-5} $

I understand the definition, but having a problem putting it into work:

1) The guess is obvious, $ \lim = 0 $: $$ \lvert x^2-2x-0 \rvert = \lvert x(x-2) \rvert = \lvert x \rvert \cdot \lvert x-2 \rvert $$

Can I finish this with saying: $$ \lvert x \rvert \cdot \lvert x-2 \rvert \leq \lvert x-2 \rvert$$ And then picked $ \delta = \epsilon $ and I'm done??

2) The guess is $ \lim = -11 $. $$ \biggr\lvert {5x+1 \over 2x-5} - ( -11 ) \biggr\rvert = \biggr\lvert {27x-54 \over 2x-5} \biggr\rvert = \biggr\lvert {27(x-2) \over 2x-5} \biggr\lvert \leq \lvert 27(x-2) \rvert = 27\lvert x- 2\rvert $$

But I'm quite lost how to continue from here..

Seems like I just need the final punch on this, thanks in advance.

Tianlalu
  • 5,177
Tegernako
  • 425

2 Answers2

1

For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|\leq 3|x-2|.$$ Then, $\delta =\min\{1, \frac{\varepsilon }{3}\}$ work. For 2), take $\delta =\frac{\varepsilon }{27}$.

user621345
  • 674
0

Actually,we need to make a form to solve $\delta$.

For 1),you have got from which $|x-2|$in it.By the definition we need to find $\delta$ with $|x-2|<\delta$ so that $|x^2-2x-0|<\varepsilon$.Now, you have

$|x||x-2|$,so need to limit the range of |x|.For example, we limit $|x-2|<h$,you can let h ever you like. Then,we have $|x|<2+h$.

So,$|x||x-2|\leq (2+h)|x-2|<\varepsilon$.Solve this inequality equation,we get $\delta_1=\frac{\varepsilon}{2+h}$.With before limited,we get $\delta=\min\{h, \frac{\varepsilon}{2+h}\}.$

For 2),you have got $27|x-2|$,rather than $g(x)|x-2|$,so you can directly solve $27|x-2|<\varepsilon$ to get $\delta=\frac{\varepsilon}{27}$.

Daeree
  • 15