Epsilon delta proof for a limit.

• Apr 22nd 2013, 05:36 PM
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

$\displaystyle \lim_{x\rightarrow 1}x^2 + 3 = 4$
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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 $\displaystyle \epsilon > 0$, there exists a $\displaystyle \delta > 0$, such that $\displaystyle |x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$:

Left hand limit:
$\displaystyle 1 - \delta < 1 \Rightarrow \delta > 0$

$\displaystyle 4 - \epsilon < 4 \Rightarrow \epsilon > 0$

Right hand limit:
$\displaystyle 1 < 1 + \delta \Rightarrow \delta > 0$

$\displaystyle 4 < 4 + \epsilon \Rightarrow \epsilon > 0$

There exists an $\displaystyle \epsilon > 0$ and a $\displaystyle \delta > 0$ 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. :/
• Apr 22nd 2013, 07:47 PM
Gusbob
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 $\displaystyle \epsilon$. I'll show you how to do it properly.

Claim: For a fixed $\displaystyle \epsilon >0$, there exists a $\displaystyle \delta>0$ such that $\displaystyle |x-1|\leq \delta \implies |x^2+3-4|\leq \epsilon$

Such a $\displaystyle \delta$, if it exists, must satisfy the following inequality:

Now $\displaystyle |x^2-1|=|(x-1)(x+1)|\leq |x-1|\cdot |x+1|\leq \delta |x+1|$

Obviously we can't actually bound $\displaystyle |x+1|$ for arbitrary $\displaystyle x$, but we don't actually care about what happens to $\displaystyle x$ far away from 1. For example, we can require $\displaystyle \delta \leq 1$. Then $\displaystyle |x-1|<\delta \leq 1 \implies 0\leq x\leq 2 \implies 1\leq |x+1| \leq 3$. For such $\displaystyle \delta$, we have

$\displaystyle |x^2-1| \leq \delta |x+1| \leq 3\delta$. If we want the last expression to be $\displaystyle \epsilon$, we require $\displaystyle \delta \leq \frac{\epsilon}{3}$.

To recap, we have
$\displaystyle |x^2-1|\leq 3\delta \leq \epsilon$

with $\displaystyle \delta$ subject to the conditions $\displaystyle \delta\leq 1$ and $\displaystyle \delta \leq \frac{\epsilon}{3}$. Choosing $\displaystyle \delta = \min\{1,\frac{\epsilon}{3}\}$ satisfies these conditions. As such you have found your desired $\displaystyle \delta$.

Writing the proof out formally:

Fix $\displaystyle \epsilon >0$ and choose $\displaystyle \delta =\min\{1,\frac{\epsilon}{3}\}$. Suppose $\displaystyle |x-1|\leq \delta$. Then

$\displaystyle |x^2+3-4|$
$\displaystyle =|x^2-1|$
$\displaystyle \leq |x-1|\cdot |x+1|$
$\displaystyle \leq \frac{\epsilon}{3} |x+1|$ since $\displaystyle |x-1| < \delta \leq \frac{\epsilon}{3}$
$\displaystyle \leq \frac{\epsilon}{3}\cdot 3$ since $\displaystyle |x-1| < \delta \leq 1 \implies |x+1|\leq 3$
$\displaystyle \leq \epsilon$
• Apr 24th 2013, 10:33 PM
hollywood
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