Show that $\lim_{x\to -3}(x^4+7x-17)=43$ using the formal definition of the limit.

Question

Tryng this by myself, I did not manage to find the solution but after looking at the correction, I don't understand an important step :

First, we suppose that there exists an $\delta>0$ such that : $0<\lvert x-(-3)\rvert\leq\delta$ $\mathbf{and}\;\delta\leq 1$. Why suppose that $\delta\leq 1$ ? This step is very important for the continuation of the correction. Intuitively it is clear that epsilon should be small, but in the formal definition of the limit delta is not required to be small, but only to depend on epsilon. So why start directly with such an assumption?

Answer 1

First, an easier example. Suppose we wanted to prove that $\displaystyle\lim_{x\to2}x^2 = 4$. Then, we would need to show that for any $\varepsilon > 0$, there is some $\delta > 0$ such that $|x^2 - 4| < \varepsilon$ whenever $0 < |x - 2| < \delta$.

The consequent may be factored: $|x^2 - 4| = |x - 2||x + 2|$, and notice that the antecedent gives us more information about $x - 2$. However, we would like more information about $x + 2$. So we stipulate that $\delta \le 1$. If this is not the case, we will deal with it later. If that is the case, then $0 < |x - 2| < 1$ or $2 < x < 3$ so $4 < x + 2 < 5$.

Now we have two pieces of information:

• $|x - 2| < \delta$, and
• $|x + 2| < 5$.

So we can say that $|x - 2||x + 2| < 5\delta$. Now if we set $\delta = \varepsilon/5$, then $|x - 2||x + 2| < \varepsilon$, and we're done if $\delta \le 1$ works. However, it may not work, if $\varepsilon > 5$. In that case, we can always choose a smaller $\delta$. So in this case, we would set $\delta = \min\{ \varepsilon/5, 1 \}$, so that even if $\varepsilon > 5$, we would instead use $\delta = 1$ and it would work.

Now for your question. We may factor the would-be consequent in the definition of the limit: $(x^4 + 7x - 17) - 43 = (x + 3)(x^3 - 3x^2 + 9x - 20)$. We will then stipulate about our $\delta$. We could stipulate that $\delta \le 5$, or $\delta \le \pi$, or even $\delta \le 10^6$ or $\delta \le 0.69420$. These values are all valid, however they make the arithmetic unnecessarily complicated, so it is easier to stipulate that $\delta \le 1$.

So stipulate that $\delta \le 1$ so $0 < |x + 3| < 1$ gives us $-3 < x < -2$. Now it is a matter of bounding the consequent. That is, we need to bound $x^3 - 3x^2 + 9x - 20$ using information about $x$ we found above. After that, we will just let $\delta$ be a minimum of $1$ and some expression involving $\varepsilon$.

To summarize, we can't say $\delta \le 1$ always, it may instead be a minimum of multiple values. To say more we would need to see the full proof.