Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition

Let $\epsilon > 0$. Then choose $\delta < \text{min}\{1,\epsilon/2\}$. If $|x-1|<\delta$, then we also have that $|x-2|> 1$ and $|x+1|<2$. Then $$ |f(x)-1| = \frac{|x-1|\cdot|x+1|}{|x-2|}< \frac{\delta|x+1|}{|x-2|}< \frac{2\delta}{|x-2|}<\frac{2\delta}{1} < \epsilon $$

$f(x) = \frac{x+1}{x-2}+x=\frac{x^2 +1-x}{x-2}$

We can prove that : $|f(x) - l|<\delta $ $ \Leftarrow $ $ |x-a|<\alpha $

$\alpha , \delta > 0$

$|f(x) - l|= |\frac{x^2 +1-x}{x-2}+1|=|\frac{(x+1)(x-1)}{x-2}|$

$|f(x) - l|<\delta$ $\Rightarrow $ $|\frac{(x+1)(x-1)}{x-2}|<\delta$

$\Rightarrow $ $|x-1| |\frac{(x+1)}{x-2}|<\delta$

Suppose $x\in [\frac{1}{2}, \frac{3}{2}] $ $\Rightarrow $

$\frac{3}{2}\leq x+1\leq\frac{5}{2}$

and:

$\frac{-3}{2} \leq x-2\leq\frac{-1}{2}$

$\Rightarrow$ $\frac{2}{3}\leq \frac{-1}{x-2}\leq2$

$\Rightarrow $ $|\frac{x+1}{x-2}|\leq5$

$\Rightarrow $ $|x-1| |\frac{(x+1)}{x-2}|\leq5|x-1|$

we know that :

$|x-1| |\frac{x+1}{x-2}|<\delta$

So:

$5|x-1|<\delta$

$\Rightarrow $ $|x-1|<\frac{\delta}{5} $

We put $\alpha=\frac{\delta} {5}$

Finally : After the definition of limite we proved $\lim_{x\to 1} f(x) =-1$