<s>[INST] Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition I got a question regarding my answer of proving limit using epsilon-delta, here's the question

Prove $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$

Here's the answer I've come up so far
let $f(x) = \frac{x+1}{x-2} + x$
by algebra manipulation, we get
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1| $ $=|\frac{x^2 - 1}{x-2}|$ $=|\frac{(x-1)(x+1)}{x-2}|$
let $|x-1| < 1$, by triangle inequality we get $|x| < 2$, then
$|x + 1| < 3$ and $|x - 2| < 1$
now, using the definiton of limit,
for every $\epsilon > 0$, there exist $\delta = min\{1, \frac{\epsilon}{3}\}$ such that
if $0 < |x - 1| < \delta$ then,
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1|$ $=|\frac{(x-1)(x+1)}{x-2}|$ $=|\frac{1 \cdot 3}{1}|$ $< \epsilon$
Is this correct? honestly I'm not sure on getting the upper bound of $|x-2|$, so I used the assumption of $|x-1| < 1$
Any tips would help, thanks beforehand.
 [/INST] $f(x) = \frac{x+1}{x-2}+x=\frac{x^2 +1-x}{x-2}$
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 $\lim_{x\to 1} f(x) =-1$
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