Given $$p\left(x\right)=x^{13}+17x^{12}-10x^{11}+1$$ Prove, using the epsilon-delta definition of limit: $$\lim _{x\to \infty }\left(p\left(x\right)^{\frac{1}{13}}-x\right)=\frac{17}{13}$$

I tried to let $x=\frac 1 y$.

I can prove it without epsilon-delta, but is there any way to prove it using epsilon-delta?

  • $\begingroup$ This is not $\epsilon$-$\delta$ approach as you are dealing with $x\to\infty$ $\endgroup$ – Juniven May 12 '17 at 10:17

You didn't mention how ou prove it without it, but I guess that you're using Taylor expansion of $(1+h)^{1/13} = 1+h/13+O(h^2)$ which gives together with $p(x) = x^{13}(1+17/x-10/x^2 +1/x^{13})$ that

$$p(x) ^ {1/13} - x = x (1 + 17/x - 10/x^2 + 1/x^{13})^{1/13} - x\\ = x ( 1 + 17/13x + O(1/x^2)) - x = 17/13 + O(1/x) \to 17/13$$

here we actually use nothing that can't be proved using epsilon, but it's the Taylor estimate part that is the harder.

Now in order to do this in the epsilon way we want to emulate the Taylor expansion. We turn to the Taylor expansion and simply try to estimate the rest term especially for $h>0$. Of course we can write

$$(1+h)^{1/13} = 1 + h/13 + f(h)$$

for some function f(h). Now we use the binomial theorem:

$$1 + h = (1+h/13+f(h))^{13} = \sum \binom{13}{k} (1+f(h))^{13-k}(h/13)^k\\ = (1+f(h))^{13} + h(1+f(h))^{12} + {h^2\over 169} \sum_2^{13}\binom{13}{k}(1+f(h))^{13-k}(h/13)^{k-2}\\ = 1 + h + \underbrace{ \sum_1^{13}\binom{13}{k}f(h)^k +\sum_1^{12}\binom{12}{k}f(k)^k + {h^2\over169} \sum_2^{13}\binom{13}{k}(1+f(h))^{13-k}(h/13)^{k-2}}_{=0}$$

Now we rewrite this as:

$$f(h)\left(\sum_1^{13}\binom{13}{k}f(h)^{k-1} +\sum_1^{12}\binom{12}{k}f(k)^{k-1}\right) = -\left(\sum_2^{13}{1\over169}\binom{13}{k}(1+f(h))^{13-k}(h/13)^{k-2}\right)h^2$$

Now we see that the expressions in the parentheses are polynomials with a non-zero constant. Since $1\le(1+h)^{1/13}\le1+h$ we have that $0\le f(h)+h/13 \le h$ which makes $-h/13\le f(h)\le 12x/13$ so certainly $|f(h)|<|h|$. So by restricting $|h|$ to be small enough we see that we can find bounds for the polynomials both above and below (but over zero). This means that we can find a constant such that

$$|f(h)| \le Cx^2$$

Now we arrived at the Taylor estimate and can then use that since if $x$ is large enuogh

$$p(x)^{1/13}-x = x(1+17/x-10/x^2+1/x^13)^{1/13} = x(1+17/13x+f(1/x))-x\\ = 17/13 + xf(1/x)$$

And by using our estimate for $f(x)$ we have:

$$C/x \ge |x f(1/x)| = \left| p(x)^{1/13}-x - {17\over13}\right|$$

So if $x$ is large enough and $x > C/\epsilon$ we have $\left| p(x)^{1/13}-x - {17\over13}\right|<\epsilon$ which proves the limit.


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