Thread: Epsilon delta proof for a limit.

1. 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

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

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

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

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

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

There exists an $\epsilon > 0$ and a $\delta > 0$ 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. :/

2. 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 $\epsilon$. I'll show you how to do it properly.

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

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

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

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

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

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

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

Writing the proof out formally:

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

$|x^2+3-4|$
$=|x^2-1|$
$\leq |x-1|\cdot |x+1|$
$\leq \frac{\epsilon}{3} |x+1|$ since $|x-1| < \delta \leq \frac{\epsilon}{3}$
$\leq \frac{\epsilon}{3}\cdot 3$ since $|x-1| < \delta \leq 1 \implies |x+1|\leq 3$
$\leq \epsilon$

3. 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