<s>[INST] Prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$ 
Using the $\epsilon,\delta$ definition of uniform continuity, prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$.

Attempt:
We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that if $|x-a|<\delta$ and $|\frac{1}{1+x}-\frac{1}{1+a}|<\epsilon$ for all $a \in (-\infty,-\frac{3}{2})$.
$\quad$Scratchwork for $\delta$:.
$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a} \Bigg|=\Bigg|\frac{1+a-1\color{blue}{-}x}{(1+x)(1+a)}\Bigg|&=\fbox{|$\frac{1}{(1+x)(1+a)}|$}|x-a|\\&\color{blue}{<}4|x-a|<\epsilon \\ &\Rightarrow|x-a|<\frac{\epsilon}{4} 
\end{align}$
$\quad \quad $ **In order to bound $\Bigg|\frac{1}{(1+x)(1+a)}\Bigg|$, I used the fact that $x,a<-\frac{3}{2}$.
$\quad \quad $Which would result in $\frac{1}{(1+x)(1+a)}<\frac{1}{(1-\frac{3}{2})(1-\frac{3}{2})}=\frac{1}{\frac{1}{4}}=4$.
$\quad$Choose $\delta=\frac{\epsilon}{4}\\$
$\quad$Proof:
Let $\epsilon>0$ be given and let $\delta=\frac{\epsilon}{4}$. If $|x-a|<\delta$ we have:
$\begin{align}
\quad \quad \Bigg|\frac{1}{1+x}-\frac{1}{1+a}\Bigg|&= \Bigg|\frac{1}{(1+x)(1+a)}\Bigg||x-a|\\&\leq4|x-a|\\ &<4\delta=4\frac{\epsilon}{4}=\epsilon. 
\end{align}$
$\rule{18cm}{.03cm}$
Is it okay to use the fact that that $x,a<-\frac{3}{2}$, in order to place bounds on the function in order to reach a $\delta$? I am not sure if this technique holds for intervals involving infinity.
 [/INST] Since $x$ and $a$ are both assumed to satisfy the inequalities $x,a < -\frac{3}{2}$ then sure, you may use those inequalities in your proof.
The proof looks fine. Your scratchwork has a mistaken inequality and a mistaken implication sign, although that seems not to have affected the proof.
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