Easy Limits Q. Homework: Show $\lim_{x\to2}\frac{x+1}{x+2}=\frac{3}{4}$

[quanticism]

Homework Statement

Show that
<br /> \lim_{x\to2}\frac{x+1}{x+2}=\frac{3}{4}.<br />

Homework Equations

Let
\epsilon&gt;0. We seek a number \delta&gt;0: if |x-2|&lt;\delta then |\frac{x+1}{x+2}-\frac{3}{4}|&lt;\epsilon.

The Attempt at a Solution

Now
<br /> |\frac{x+1}{x+2}-\frac{3}{4}|=|\frac{4x+4-3x-6}{4(x+2)}|=|\frac{x-2}{4(x+2)}|.<br />

So for
<br /> |\frac{x+1}{x+2}-\frac{3}{4}|&lt;\epsilon, we require |\frac{x-2}{4(x+2)}|&lt;\epsilon.<br />

ie.
<br /> |\frac{x-2}{x+2}|&lt;4\epsilon.<br />

I got stuck here since I'm not sure how to express
<br /> |\frac{x-2}{x+2}|<br />
in terms of |x-2|

 

[Berko]

So, |x-2| < 4e|x+2| = delta...

 

[quanticism]

So, |x-2| < 4e|x+2| = delta...

Don't we have to express delta as a function of epsilon only?

[strike]Or am I meant to argue that since we're looking in the neighbourhood around x=2, we can choose delta to be <4e|4|=16e. But this statement doesn't really click with me so if it's true, can someone clarify why this works[/strike]

Edit: Pretty sure that ^ isn't going down the right track.

 

[praharmitra]

You are supposed to get \delta in terms of \epsilon only. Try this write x+2 = x-2+4, and divide both the numerator and denominator by (x-2). Now the entire expression contains only (x-2). Now can you simplify this further?.

Hint: use |a|+|b| >= |a+b|

 

[quanticism]

Oh wow. That certainly worked out nicely :) I got
|x-2|&lt;\frac{4\epsilon}{1-\epsilon}.

But would that mean we epsilon needs to be chosen so that 0&lt;\epsilon&lt;1? But this is a limits question so I guess we're only really concerned about when \epsilon is small.

 

[praharmitra]

Oh wow. That certainly worked out nicely :) I got
|x-2|&lt;\frac{4\epsilon}{1-\epsilon}.

But would that mean we epsilon needs to be chosen so that 0&lt;\epsilon&lt;1? But this is a limits question so I guess we're only really concerned about when \epsilon is small.

Can you recheck your calculations? Because I got,

|x-2} &lt; \frac{16\epsilon}{1-4\epsilon}

with the way you had defined the problem. Of course its only a matter of redefinition of \epsilon, but it is always advisable to use the same notation everywhere.

And yes, since we are working with limits, we are only concerned about when \epsilon is small

 

[HallsofIvy]

|x- 2|&lt; \frac{\epsilon}{|x+ 2|}

I would have been inclined to say:

If |x- 2|< 1, then -1< x- 2< 1 so that 3< x+ 2< 5 and |x+ 2|< 5 That means that
\frac{1}{5}&lt; \frac{1}{|x+ 2|}
so that
\frac{\epsilon}{5}&lt; \frac{\epsilon}{|x+ 2|}

That is, as long as |x- 2|< 1, we can take
\delta= \frac{\epsilon}{5}[/itex]<br /> If |x- 2|&lt; \delta, then it is also less than \epsilon/5&amp;lt; \epsilon/|x+2| and, working backwards, <br /> \left|\frac{x+1}{x+ 2}- \frac{3}{4}\right|&amp;lt; \epsilon<br /> as you wanted.<br /> <br /> and then, working backwards,<br /> \left|\frac{x+ 1}{x+ 2}- \frac{3}{4}\right|&amp;lt; \epsilon.<br /> <br /> Of course that &quot;|x- 2|&lt; 1&quot; is also important so we would have to take \delta to be the smaller of \epsilon/5 and 1 so that those are both true.

 

[quanticism]

I'll show what I did and you can point out any errors/invalid algebraic manipulations.

|\frac{x-2}{x+2}|&lt;\epsilon ...(1)
|\frac{x-2}{x-2+4}| &lt; \epsilon
|\frac{1}{1+\frac{4}{x-2}} &lt;\epsilon
Now note that by applying triangle inequality:

\frac{1}{1+|\frac{4}{x-2}|}} \le \frac{1}{|1+\frac{4}{x-2}}|}

So if (1) is true, then
\frac{1}{1+|\frac{4}{x-2}|} &lt;\epsilon
1+ |\frac{4}{x-2}| &gt;1/\epsilon (valid step since both sides are positive)
\frac{4}{|x-2|} &gt;\frac{1-\epsilon}{\epsilon}
|x-2|&lt;\frac{4\epsilon}{1-\epsilon} (once again, both sides were positive)

Typing in latex is rather time consuming. I should probably just use paint and my tablet.

 

[quanticism]

|x- 2|&lt; \frac{\epsilon}{|x+ 2|}

Shouldn't it be |x-2|<e|x+2| ?

But apart from that I think I see where you're going.

If |x-2|<1, then |x+2|<5.

So we have |x-2|<5e<e|x+2| ...(1)

So we pick a \delta to be 5e such that 0<\delta<1 which would work if epsilon was small.

Edit: Opps, (1) should be |x-2|<e|x+2|<5e which isn't really useful.

Edit2: If |x-2|<1, then 3<x+2<5 so 3<|x+2|<5

So we have |x-2|<3e<e|x+2|. Now we can choose delta to be 3e? Can't say I'm too convinced with what I did here though.

 

[praharmitra]

I'll show what I did and you can point out any errors/invalid algebraic manipulations.

|\frac{x-2}{x+2}|&lt;\epsilon ...(1)
|\frac{x-2}{x-2+4}| &lt; \epsilon
|\frac{1}{1+\frac{4}{x-2}} &lt;\epsilon
Now note that by applying triangle inequality:

\frac{1}{1+|\frac{4}{x-2}|}} \le \frac{1}{|1+\frac{4}{x-2}}|}

So if (1) is true, then
\frac{1}{1+|\frac{4}{x-2}|} &lt;\epsilon
1+ |\frac{4}{x-2}| &gt;1/\epsilon (valid step since both sides are positive)
\frac{4}{|x-2|} &gt;\frac{1-\epsilon}{\epsilon}
|x-2|&lt;\frac{4\epsilon}{1-\epsilon} (once again, both sides were positive)

Typing in latex is rather time consuming. I should probably just use paint and my tablet.

Your equation (1) should be

|\frac{x-2}{x+2}|&lt;4\epsilon

check what you wrote in your first post.

 

[quanticism]

Your equation (1) should be

|\frac{x-2}{x+2}|&lt;4\epsilon

check what you wrote in your first post.

Ah sorry, you are correct. Guess I was so focused on the x's and epsilons that I forgot about the constant.