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

[quanticism]

Homework Statement

Show that
$$\lim_{x\to2}\frac{x+1}{x+2}=\frac{3}{4}.$$

Homework Equations

Let
$$\epsilon>0$$. We seek a number $$\delta>0$$: if $$|x-2|<\delta$$ then $$|\frac{x+1}{x+2}-\frac{3}{4}|<\epsilon.$$

The Attempt at a Solution

Now
$$|\frac{x+1}{x+2}-\frac{3}{4}|=|\frac{4x+4-3x-6}{4(x+2)}|=|\frac{x-2}{4(x+2)}|.$$

So for
$$|\frac{x+1}{x+2}-\frac{3}{4}|<\epsilon$$, we require $$|\frac{x-2}{4(x+2)}|<\epsilon.$$

ie.
$$|\frac{x-2}{x+2}|<4\epsilon.$$

I got stuck here since I'm not sure how to express
$$|\frac{x-2}{x+2}|$$
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
$$|x-2|<\frac{4\epsilon}{1-\epsilon}.$$

But would that mean we epsilon needs to be chosen so that $$0<\epsilon<1$$? 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
$$|x-2|<\frac{4\epsilon}{1-\epsilon}.$$

But would that mean we epsilon needs to be chosen so that $$0<\epsilon<1$$? 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,

$$|x-2} < \frac{16\epsilon}{1-4\epsilon}$$

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]

$$|x- 2|< \frac{\epsilon}{|x+ 2|}$$

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
$$\frac{1}{5}< \frac{1}{|x+ 2|}$$
so that
$$\frac{\epsilon}{5}< \frac{\epsilon}{|x+ 2|}$$

That is, as long as |x- 2|< 1, we can take
[tex]\delta= \frac{\epsilon}{5}[/itex]<br /> If |x- 2|< \delta, then it is also less than $\epsilon/5< \epsilon/|x+2|$ and, working backwards, <br /> $$\left|\frac{x+1}{x+ 2}- \frac{3}{4}\right|< \epsilon$$<br /> as you wanted.<br /> <br /> and then, working backwards,<br /> $$\left|\frac{x+ 1}{x+ 2}- \frac{3}{4}\right|< \epsilon$$.<br /> <br /> Of course that "|x- 2|< 1" 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.[/tex]

 

[quanticism]

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

$$|\frac{x-2}{x+2}|<\epsilon$$ ...(1)
$$|\frac{x-2}{x-2+4}| < \epsilon$$
$$|\frac{1}{1+\frac{4}{x-2}} <\epsilon$$
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
$$\frac{1}{1+|\frac{4}{x-2}|} <\epsilon$$
$$1+ |\frac{4}{x-2}| >1/\epsilon$$ (valid step since both sides are positive)
$$\frac{4}{|x-2|} >\frac{1-\epsilon}{\epsilon}$$
$$|x-2|<\frac{4\epsilon}{1-\epsilon}$$ (once again, both sides were positive)

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

 

[quanticism]

$$|x- 2|< \frac{\epsilon}{|x+ 2|}$$

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.

$$|\frac{x-2}{x+2}|<\epsilon$$ ...(1)
$$|\frac{x-2}{x-2+4}| < \epsilon$$
$$|\frac{1}{1+\frac{4}{x-2}} <\epsilon$$
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
$$\frac{1}{1+|\frac{4}{x-2}|} <\epsilon$$
$$1+ |\frac{4}{x-2}| >1/\epsilon$$ (valid step since both sides are positive)
$$\frac{4}{|x-2|} >\frac{1-\epsilon}{\epsilon}$$
$$|x-2|<\frac{4\epsilon}{1-\epsilon}$$ (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

$$|\frac{x-2}{x+2}|<4\epsilon$$

check what you wrote in your first post.

 

[quanticism]

Your equation (1) should be

$$|\frac{x-2}{x+2}|<4\epsilon$$

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.