Prove limit by definition $\lim _{x\to 3}\left(\frac{2-x}{x+1}\right)=-\frac{1}{4}$

Question

Prove by the definition that $\lim _{x\to 3}\left(\frac{2-x}{x+1}\right)=-\frac{1}{4}$

I started like that: $$\lim _{x\to 3}\left(\frac{2-x}{x+1}\right)=-\frac{1}{4} \longrightarrow \left|\frac{2-x}{x+1}+\frac{1}{4}\right|< \epsilon$$ Then $$= \left|\frac{-3x+9}{4\left(x+1\right)}\right|<\epsilon = \:\frac{3}{4}\left|\frac{-x+3}{x+1}\right| < \epsilon$$ Now i solved the system: $$\:\frac{-x+3}{x+1}<\frac{4\epsilon}{3} \longrightarrow (\frac{4\epsilon}{3}-1-\frac{4\epsilon}{3} \longrightarrow (-\frac{4\epsilon}{3}-1

obviously the result is not correct because in neither intervals i found there is the 3 $(x \to 3)$. Can you give me a hand to set the following problem? I can not understand where i'm wrong

Answer 1

What you want is to see that, for any $\varepsilon>0$, the solutions of $$ \left|\frac{2-x}{x+1}+\frac{1}{4}\right|< \varepsilon $$ include a punctured neighborhood of $3$. Your transformation is good: $$ \left|\frac{3-x}{x+1}\right|<\frac{4}{3}\varepsilon $$ Now, if $x>2$, you have $x+1>3$, so $(x+1)^{-1}<\frac{1}{3}$.

So if we take $x\in(3-\delta,3+\delta)$, with $\delta<1$, we have $$ |3-x|<\delta $$ and therefore $$ \left|\frac{3-x}{x+1}\right|<\frac{1}{3}|3-x|<\frac{\delta}{3} $$ Choose $\delta=\min\{4\varepsilon,1\}$.

Answer 2

The part after "Now I solved the system:" is incorrect, I'm not sure what you did there to comment on what exactly when wrong.

But let's stop at the last thing you did before, you transformed the $\epsilon$-notation of the limit into

$$ \frac{3}{4}\left|\frac{-x+3}{x+1}\right| < \epsilon$$

The limit definition new requires you to find a $\delta$ such that the above holds true if $\left|x-3\right| < \delta$. The enumerator of the fraction is a term we already know something about: $\left|-x+3\right| = \left|x-3\right| < \delta$. The denominator $\left|x+1\right|$ "will be near 4 when $x$ is near 3". To get that more formal, let's assume $\delta \le 1$, that means $2 \le x \le 4$, hence $x+1 \ge 3$ and thus $\left|\frac{1}{x+1}\right| \le \frac{1}{3}$. Thus your requirements transforms to

$$ \frac{3}{4}\delta\frac{1}{3} < \epsilon$$

which yields to $\delta < 4\epsilon$.

So now we can finally prove the limit by going in reverse:

Set

$$\delta = \min (4\epsilon, 1)$$

Then, under the assumption $\left|x-3\right| < \delta$, we get

$ \frac{3}{4}\left|\frac{-x+3}{x+1}\right| = \frac{3}{4}\left|-x+3\right|\frac{1}{\left|x+1\right|} < \frac{3}{4}\delta\frac{1}{3} = \frac{\delta}{4} \le \epsilon$,

which is what we needed to prove.