Epsilon-Delta definition of Limit #2

Brief Explanation: Today we continue our study of epsilon-delta proofs. If you haven’t seen our previous exercise on this topic, check it here.

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Question: Prove the following limit using the Epsilon-Delta definition of limit:

enunciado

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Comprehending: The first step is always writing the following statement:

For every ε > 0 there is some δ > 0 such that for all x,

if 0 < |x – a|< δ then |f(x) – l|< ε.

Applying to our case, we have:

1.png

Next, we must relate the expressions we have above. This means trying to relate the epsilon inequality into the delta inequality. In this case, I decided to do something like this:

2.png

When the terms inside the absolute value bars are multiplying, we know we can take them out individually:

3.png

So, we have:

4.png

We now have the |x-a| that we see in the delta expression! Given that we have many terms multiplying, we’ll have to bound our delta. Let’s set δ = 1.

5.png

Using the basic inequality |x|-|a|<|x – a|:

6.png

(notice that we ignored the |x – a| from the first to the second step)

Now we have a bound for |x|! We’ll use this to bound the expressions |x^2+a^2| and |x+a|. Let’s start with |x+a|and remember the inequality |x+a|<|x|+|a|:

7.png

So we know |x+a| < 1 + 2|a|! That was an easy expression to bound. Now, for the next one:

8.png

Using our x bound:

9.png

To sum up, we got the following bounds:

10.png

Returning to our epsilon expression:

4.png

11.png

12.png

And that’s it! We have our bound for |x-a| and consequently we have our delta values! Remember that to get this result we had to set δ = 1 in the beginning of the exercise. This will mean that:

13.png

And we’re done! Sorry if the expressions are a bit long, if you have any doubts don’t hesitate to contact me!

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2 thoughts on “Epsilon-Delta definition of Limit #2

  1. Pingback: Epsilon-Delta definition of Limit #3 – Matiphy

  2. Pingback: Epsilon-Delta definition of Limit #4 – Matiphy

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