# 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!

$$\lim_{x\to 3} \frac{2}{x+1} =\frac12$$

• That is supposed to say lim as x approaches 3 – Jacob Culleny Sep 16 '14 at 20:36
• Isn't there any example that looks familiar? How about obtaining an expression for $\frac{2}{x+1} - \frac{1}{2}$? – imranfat Sep 16 '14 at 20:37
• Its how my teacher gave it to us. It just means the limit value is 3. Heres the step im stuck at: f(x)=L+- ε 2=(1/2+-ε)x + 1 x= 3+- 2/ε I think i have to define values for x(0) and x(1) – Jacob Culleny Sep 16 '14 at 20:44
• check this out: math.stackexchange.com/questions/65667/… – imranfat Sep 16 '14 at 20:54
• You also might find this helpful: math.stackexchange.com/questions/930576/… – user84413 Sep 16 '14 at 21:20

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

$\displaystyle0<|x-3|<\delta\implies\frac{|x-3|}{2|x+1|}=\frac{|x-3|}{2}\frac{1}{|x+1|}<\epsilon$.

• Thank you so much for taking the time to help me!! – Jacob Culleny Sep 17 '14 at 1:54
• You're welcome. (The actual proof would start something like this: Let $\epsilon>0$ be given, and let $\delta=\cdots$.) – user84413 Sep 17 '14 at 20:22

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.