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!

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

  • $\begingroup$ That is supposed to say lim as x approaches 3 $\endgroup$ – Jacob Culleny Sep 16 '14 at 20:36
  • $\begingroup$ Isn't there any example that looks familiar? How about obtaining an expression for $\frac{2}{x+1} - \frac{1}{2}$? $\endgroup$ – imranfat Sep 16 '14 at 20:37
  • $\begingroup$ 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) $\endgroup$ – Jacob Culleny Sep 16 '14 at 20:44
  • $\begingroup$ check this out: math.stackexchange.com/questions/65667/… $\endgroup$ – imranfat Sep 16 '14 at 20:54
  • $\begingroup$ You also might find this helpful: math.stackexchange.com/questions/930576/… $\endgroup$ – user84413 Sep 16 '14 at 21:20


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


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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.