Epsilon-Delta definition of Limit #3

Brief Explanation: Today we continue our study of epsilon-delta proofs. If you haven’t seen our previous exercises on this topic, check it here (#1) and here (#2).

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

enunciado.png

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Comprehending: I decided to post this limit just because it’s a rational function. As you’ll see, it’s quite easy but it requires some different solving technique. Let’s write the general statement (again, sorry!) regarding these types of proofs:

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

if 0 < |x – a|< δ then |f(x) – l|< ε.

For our specific case, this will mean (a = 1, l = 1):

1.png

Now, let’s relate the delta and the epsilon expression. Developing the second we get:

2.1.png

Remember that inside the absolute bars we have |a-x|=|-(a-x)|=|x-a|:

2.2.png

So the inequality becomes:

2.3.png

What do we do in these situations? The trick is to to bound |x-1| and then evaluate the bound that gives on our |1/x|. This will be possible if we assume a value for δ:

2.png

(we used δ=1/2 because if we chose δ=1 we’d run into a problem as you’ll see in the end)

Using the using the inequality to get rid of the absolute value bars:

3.png

4.png

5.png

Returning to our epsilon condition and applying this bound on 1/x:

7.png

8.png

And that’s it! Remember we had to bound the |x-1|, so our delta is:

9.png
And we’re done!

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One thought on “Epsilon-Delta definition of Limit #3

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

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