Proving $\lim_{x\rightarrow 1} (x^2+3x-1)=1$ using $\delta-\epsilon$; Why is $\delta \lt \frac{\epsilon}{6}$ correct?

Question

Using $\delta-\epsilon$ I find that using the solution manuals $\delta \lt 1$:

$$|x+4|x-1|\lt6|x-1|\lt \epsilon$$

This leave me with:

$$|x-1|\lt \frac{\epsilon}{6} $$

Now I can pick that $\delta\lt$ min{1,$\frac{\epsilon}{6}$} and this satisfies the condition on |x-1|, correct?

My questions are:

1.) Why have we picked $\delta \lt 1$ and not $\delta=1$ or any other number for that matter?

2.) When is picking a certain delta wrong? (i.e. maybe when there's an issue with the domain?)

3.) How do I do a post-proof check to make sure what I did is correct?

Answer 1

The reason they picked $\delta < 1$ is so that: $$\lvert x-1 \rvert < \delta < 1 \implies -1 < x-1 < 1 \implies 0 < x < 2$$ Thus, $x$ is between $0$ and $2$ so $\lvert x+4 \rvert$ is now bounded between $4$ and $6$ and thus $\lvert x+4 \rvert < 6$. We can now divide both sides of the inequality by $6$ as they did to bound $\lvert x-1 \rvert$ in terms of $\epsilon$.

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However, we didn't need to pick $1$. Let's pick $e^\pi$ instead because why not? If $\delta < e^\pi$, then we have: $$\lvert x-1 \rvert < \delta < e^\pi \implies -e^\pi < x-1 < e^\pi \implies -e^\pi+1 < x < e^\pi+1$$ Now, $\lvert x+4 \rvert$ has a maximum of either $\lvert -e^\pi+1+4 \rvert=\lvert 5-e^\pi \rvert$ or a maximum of $\lvert e^\pi+1+4 \rvert=\lvert e^\pi+5 \rvert$. It doesn't matter which one is bigger; what matters is that $\lvert x+4 \rvert$ is bounded by a maximum which we'll call $m$. Thus, we have: $$\lvert x+4 \rvert\cdot \lvert x-1 \rvert < m\lvert x-1 \rvert < \epsilon \implies \lvert x-1 \rvert < \frac{\epsilon}{m}$$ Thus, we have $\delta=\min\left(e^\pi, \frac{\epsilon}{m}\right)$ where $m=\max(\lvert 5-e^\pi\rvert, \lvert e^\pi+5\rvert)$.

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Now, to do a post-proof check, we start with our inequality $\lvert x-1 \rvert < \min(1, \frac{\epsilon}{6})$ and then work our way toward $\lvert x^2+3x-1 \rvert < \epsilon$. Now, the nice thing about our initial inequality is that we really have two inequalities: $$\lvert x-1 \rvert < \min\left(1, \frac{\epsilon}{6}\right) < 1$$ $$\lvert x-1 \rvert < \min\left(1, \frac{\epsilon}{6}\right) < \frac{\epsilon}{6}$$ Now, by the first inequality, we know $\lvert x+4 \rvert$ is bounded by $6$. Therefore, we get: $$\lvert x^2+3x-1 \rvert \leq \lvert x+4 \rvert \cdot \lvert x-1 \rvert < 6\cdot \frac{\epsilon}{6} < \epsilon$$ Notice how our post-proof check is very similar to our proof, but this time we start with the answer instead of working backwards to end with the answer. This makes sure that nothing we did going backwards can be reversed to go forwards.

Answer 2

It is easier to put $x=1+\delta$ (where $\delta$ may be positive or negative). Substituting in $f(x)=x^2+3x-1$ we get $f(1+\delta)=1+5\delta+\delta^2$. We want this to be in $(1-\epsilon,1+\epsilon)$.

The reason for taking $\delta<1$ is that we then have $0<\delta^2<|\delta|$. So we can say that $|5\delta+\delta^2|<6|\delta|$. So it is clear that this will give $|f(1+\delta)-1|<\epsilon$ provided also $|\delta|<\frac{\epsilon}{6}$.

I hope that answers 1). On 2) remember that $\epsilon$ can have any positive value. Any $\delta$ is correct provided that it works! It must give $|f(x)-1|<\epsilon$ for $|x-1|<\delta$.

On 3) post-proof checks are only really needed when things are complicated. Then it is good practice to see if your solutions works for one or two simple values.