Epsilon delta proof for a limit.

Just had an exam where the question was roughly:

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Using an epsilon delta definition of limit, prove that the limit

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I am not very confident with proving the limit definition of quadratics..

So I folded and decided to do it by two sided limit.

Rough attempt

For an , there exists a , such that :

Left hand limit:

Right hand limit:

There exists an and a on both sides of 1, which approach the value of 4.

Thefore the limit does exist, and it is 4. ▫

Pretty sure I messed up monumentally.. Linear functions are so much easier to prove than quadratics. :/

Re: Epsilon delta proof for a limit.

Erm...I'm not sure what you're doing here. It seems to be a bit of nonsense sorry - you don't get to choose . I'll show you how to do it properly.

Claim: For a __fixed__ , there exists a such that

Such a , if it exists, must satisfy the following inequality:

Now

Obviously we can't actually bound for arbitrary , but we don't actually care about what happens to far away from 1. For example, we can require . Then . For such , we have

. If we want the last expression to be , we require .

To recap, we have

with subject to the conditions and . Choosing satisfies these conditions. As such you have found your desired .

Writing the proof out formally:

Fix and choose . Suppose . Then

since

since

Re: Epsilon delta proof for a limit.

There's a good explanation of how to write these epsilon-delta proofs - it's one of the four "sticky" threads on the Calculus forum here at MHF.

- Hollywood