# This kinda special limit

Let $P_a$ and $P_b$ be 2 polynomials with same degree, n, and same leading coefficient c, with $c\neq 0$ (otherwise the degree is less than n):

$P_a(x)=cx^n+a_{n-1}x^{n-1}+\displaystyle\sum_{i=0}^{n-2}a_ix^i$

$P_b(x)=cx^n+b_{n-1}x^{n-1}+\displaystyle\sum_{i=0}^{n-2}b_ix^i$

Let $y=\displaystyle\lim_{x\rightarrow\infty}\Big(\sqrt[n]{P_a(x)}-\sqrt[n]{P_b(x)}\Big)$. In this note I try to find the value of $y$.

Let $L_a(x)=\displaystyle\sum_{i=0}^{n-2}a_ix^i$, and $L_b(x)=\displaystyle\sum_{i=0}^{n-2}b_ix^i$. Later, we'll see that both the value of $L_a(x)$ and $L_b(x)$ won't affect the value of the limit. We have:

$y=\displaystyle\lim_{x\rightarrow\infty}\sqrt[n]{c}\Bigg(\sqrt[n]{x^n+\dfrac{a_{n-1}}{c}x^{n-1}+\dfrac{L_a(x)}{c}}-\sqrt[n]{x^n+\dfrac{b_{n-1}}{c}x^{n-1}+\dfrac{L_b(x)}{c}}\Bigg)$

$y=\sqrt[n]{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{\bigg(x^n+\dfrac{a_{n-1}}{c}x^{n-1}+\dfrac{L_a(x)}{c}\bigg)-\bigg(x^n+\dfrac{b_{n-1}}{c}x^{n-1}+\dfrac{L_b(x)}{c}\bigg)}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\frac{P_a(x)}{c}}\Bigg)^i\Bigg(\sqrt[n]{\frac{P_b(x)}{c}}\Bigg)^{n-1-i}}$

$y=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{(a_{n-1}-b_{n-1})x^{n-1}+\big(L_a(x)-L_b(x)\big)}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\dfrac{P_a(x)}{c}}\Bigg)^i\Bigg(\sqrt[n]{\dfrac{P_b(x)}{c}}\Bigg)^{n-1-i}}$

$y=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\dfrac{a_{n-1}-b_{n-1}+\dfrac{L_a(x)-L_b(x)}{x^{n-1}}}{\displaystyle\sum_{i=0}^{n-1}\Bigg(\sqrt[n]{\dfrac{P_a(x)}{cx^n}}\Bigg)^i\Bigg(\sqrt[n]{\dfrac{P_b(x)}{cx^n}}\Bigg)^{n-1-i}}$

Since $\deg(P_a)=\deg(P_b)=n$ and both have leading coefficient c, then:

$y=\dfrac{\sqrt[n]{c}}{c}\displaystyle\lim_{x\rightarrow\infty}\frac{a_{n-1}-b_{n-1}}{\displaystyle\sum_{i=0}^{n-1}1^{i}1^{n-1-i}}$

$y=\dfrac{\sqrt[n]{c}}{c}\dfrac{a_{n-1}-b_{n-1}}{n}$

$y=\dfrac{a_{n-1}-b_{n-1}}{n\sqrt[n]{c^{n-1}}}$

and we're done. We've found the value of $y$. Take quite long, around 1 hour.

Note by Gian Sanjaya
4 years, 6 months ago

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You should state what you are trying to achieve at the start, so that others can understand what you are trying to say.

Staff - 4 years, 6 months ago

Oh okay, thanks for the suggestion. I'll edit it.

- 4 years, 6 months ago

Thanks. This clarifies the result :)

What happens if $\deg P_a = n$ and $\deg P_b = m$? What would be the conditions under which $\lim \sqrt[n]{P_a} - \sqrt[m]{P_b}$ is a constant? And under these conditions, what is the constant equal to?

Staff - 4 years, 6 months ago

Let the leading coefficient of $P_a$ and $P_b$ be $z_a$ and $z_b$, respectively. I would consider separating the root of polynomials in the limit to get:

$\displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[m]{P_b(x)}\Big)=\displaystyle \lim_{x\rightarrow \infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[n]{z_a}x\Big) - \displaystyle \lim_{x\rightarrow \infty} \Big(\sqrt[m]{P_b(x)} - \sqrt[m]{z_b}x\Big) + \displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{z_a} - \sqrt[m]{z_b}\Big)x$

Applying what I've found, we would see that the first 2 limits are finite; the whole limit will be constant, not approaching either sides of infinity, if and only if the last limit is constant, which happens if and only if $\sqrt[n]{z_a}-\sqrt[m]{z_b}=0$, or equivalently $\sqrt[n]{z_a}=\sqrt[m]{z_b}$. Now, let $z_a = c^n$ and $z_b=c^m$ for a positive real number c (so it holds for all values of m and n), and let $P_a$ and $P_b$ be defined as:

$P_a(x)=c^n x^n + a_{n-1} x^{n-1} + \displaystyle \sum_{i=0}^{n-2} a_i x^i$

$P_b(x)=c^m x^m + b_{m-1} x^{m-1} + \displaystyle \sum_{i=0}^{m-2} b_i x^i$

We now have:

$\displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[m]{P_b(x)}\Big)=\frac{a_{n-1}}{nc^{n-1}}-\frac{b_{m-1}}{mc^{m-1}}$

- 4 years, 6 months ago

That's a good analysis.

Pointers
1. Your end result disagrees with the previous result.
2. When you split out a limit (LHS) into 2 or more parts (RHS), it is not true that the LHS exists if and only if the RHS exists. If the RHS exists, then the LHS exists. But it is possible for the LHS to exist, while the RHS does not exist. Case in point:
$\lim 0 \neq \lim x + \lim (-x).$
3. I suggest doing something similar to what you did. Consider $( a^{nm} - b^{nm}) = (a-b) \times$ stuff.

Staff - 4 years, 6 months ago