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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!

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

Martin Sleziak
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Jacob Culleny
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  • 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: http://math.stackexchange.com/questions/65667/how-to-prove-a-limit-exists-using-the-epsilon-delta-definition-of-a-limit?rq=1 – imranfat Sep 16 '14 at 20:54
  • You also might find this helpful: http://math.stackexchange.com/questions/930576/calculus-limit-epsilon-delta – user84413 Sep 16 '14 at 21:20
  • I just really need help setting it up for this kind of problem. – Jacob Culleny Sep 16 '14 at 21:25
  • The last link you sent me was the last example i did in class. I fully understand how it works but when ever i get a new problem i just dont get how to start it. I created values for delta (o) and delta (1).I got Delta (0) equals 2/ε and for delta (1) i got -2/ε. However, i think that is incorrect. – Jacob Culleny Sep 16 '14 at 21:53
  • I don't understand what you mean by "f(x)=L+- ε 2=(1/2+-ε)x + 1 x= 3+- 2/ε". can you state in your post using the $\varepsilon$-$\delta$ notation what you have to prove? – miracle173 Sep 16 '14 at 22:05

2 Answers2

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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

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

user84413
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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.

Divide1918
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