# Proving function limit using $\epsilon$-$\delta$ definition

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.

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 $$\delta =\frac{\varepsilon }{27}$$.
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 $$|x-2|,you can let h ever you like. Then,we have $$|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}$$.