Proving a limit with epsilon delta definition: $\lim\limits_{x\to 3} \frac{2}{x+1} =\frac12$ I am in honors Calculus I and my teacher is really stressing this limit proof. I understand the examples she goes over in class but she gave us a problem for home work and i just don't know how to start it. I appreciate any help!
$$\lim_{x\to 3}  \frac{2}{x+1} =\frac12$$ 
 A: Hint:
To get started, you need to work with the inequality $\displaystyle\left|\frac{2}{x+1}-\frac{1}{2}\right|<\epsilon$.  We can rewrite this as $\displaystyle\left|\frac{4-(x+1)}{2(x+1)}\right|<\epsilon$, which gives $\displaystyle\left|\frac{3-x}{2(x+1)}\right|<\epsilon$ or, equivalently, $\displaystyle\frac{\left|x-3\right|}{2|x+1|}<\epsilon$.
Now we need to get an upper bound for the factor $\frac{1}{|x+1|}$, so one way to do this is to assume 
that our value of $\delta \le 1$.  Under this assumption, 
$0<|x-3|<\delta\implies|x-3|<1\implies 2<x<4\implies3<x+1<5\implies$
$\;\;\;\displaystyle\frac{1}{3}>\frac{1}{x+1}>\frac{1}{5}\implies \frac{1}{\left|x+1\right|}<\frac{1}{3}$.

Now you need to find a $\delta>0$ which satisfies $\delta\le1$,  and
$\displaystyle0<|x-3|<\delta\implies\frac{|x-3|}{2|x+1|}=\frac{|x-3|}{2}\frac{1}{|x+1|}<\epsilon$.
A: Given $ \delta \gt 0 $ such that $|x-3|\lt \delta, x\in(3-\delta,3+\delta)\subset[2,4]$,  $$ |\dfrac{2}{x+1} -\dfrac12 |=|\dfrac{3-x}{2(x+1)}|\lt \dfrac{\delta}{2|x+1|}\lt \dfrac{\delta}{8}:= \varepsilon$$. 
Hence, $ \forall \varepsilon \gt 0, \exists \delta \gt 0 $ such that $|x-3|\lt \delta$ implies  $ |\dfrac{2}{x+1} -\dfrac12 | \lt \varepsilon$. By definition, $\lim_{x\to 3}  \dfrac{2}{x+1} =\dfrac12$. This proof also shows the given function is uniformly continuous at x=3.
