<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] Note that by definition, let $f(x):=\frac{x+1}{x-2}+x$ so, we need to prove that
\begin{eqnarray}
\boxed{\lim_{x\to 1} f(x)=-1 \iff \forall \epsilon>0, \exists \delta>0; \forall x: 0<|x-1|<\delta \implies |f(x)-(-1)|<\epsilon}
\end{eqnarray}
Now, we need to analyse (this part is not the formal prove) the $(*)$ here:
\begin{eqnarray}
\lim_{x\to 1} f(x)=-1 \iff \forall \epsilon>0, \exists \delta>0; \forall x: 0<|x-1|<\delta \implies \underbrace{|f(x)-(-1)|<\epsilon}_{*}
\end{eqnarray}
Note that
\begin{eqnarray}
|f(x)-(-1)|&=&\left|\frac{x+1}{x-2}+x-(-1) \right|\\
&=&\left|\frac{x+1}{x-2}+x+1 \right|\\
&=& \left|\frac{x+1}{x-2}+(x+1)\left(\frac{x-2}{x-2}\right) \right|\\
&=& \left|\frac{x+1+(x+1)(x-2)}{x-2} \right|\\
&=&\left|\frac{x+1+x^{2}-2x+x-2}{x-2} \right|\\
&=&\left|\frac{x^{2}-1}{x-2} \right|\\
&=&\left|\frac{(x-1)(x+1)}{x-2} \right|\\
&=&\left|x-1\right|\cdot\left|\frac{x+1}{x-2}\right|
\end{eqnarray}
Now, we want that:
\begin{eqnarray}
\text{si} \quad 0<|x-1|<\delta \implies \left|x-1\right|\cdot \left| \frac{x+1}{x-2}\right|<\epsilon
\end{eqnarray}
Now, note that if we can find a constant $C$ such that $\left|\frac{x+1}{x-2}\right|<C$, so we have $$\left|x-1\right|\cdot \left|\frac{x+1}{x-2}\right|<C|x-1|$$
and we can do $C|x-1|<\epsilon$ taking $|x-1|<\epsilon/C=\delta$
We can find a number $C$ such that we restrict $x$ to some interval centered on $1$.
In fact, we are interested only in values close to $1$, so it's reasonable to assume that $x$ is within a distance $1$ of $1$, that is $|x-1|<1$. So, we can see that $$|x+1|=|x+1-1+1|\leq |x-1|+|2|=1+2=3 \implies |x+1|\leq 3$$
and using reverse triangle inequality and since $|x-1|>0$, we have $$|x-2|=|(x-1)-1|\geq ||x-1|-|1||>|0-1|=|-1|=1 \implies |x-2|>1$$
Note that $$|f(x)-(-1)|=\left|x-1\right|\left| \frac{x+1}{x-2}\right|<\delta\cdot \frac{3}{1}<\epsilon$$
To ensure that all inequalities are satisfied, we take $ \delta $ as the smallest of the numbers $1$ and $\epsilon/3$. The notation for this is $\delta = \min \{1, \epsilon/ 3\}$

Can you continue from here?
You need to write the formal prove.
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