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

Question

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.

Answer 1

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 $\delta = \min \{1, \epsilon/ 3\}$

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

Answer 2

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 $$ |f(x)-1| = \frac{|x-1|\cdot|x+1|}{|x-2|}< \frac{\delta|x+1|}{|x-2|}< \frac{2\delta}{|x-2|}<\frac{2\delta}{1} < \epsilon $$

Answer 3

Let's take $\delta<\frac{1}{2}$, then $|x-1|<\delta$ gives $|x-2|>\frac{1}{2}$ and $|x+1|=x+1<\frac{5}{2}$, so we have $$\left|\frac{(x-1)(x+1)}{x-2}\right|<5\delta <\varepsilon$$

Answer 4

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

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 $\lim_{x\to 1} f(x) =-1$