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.

## Comments

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TopNewestYou should state what you are trying to achieve at the start, so that others can understand what you are trying to say. – Calvin Lin Staff · 1 year, 11 months ago

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– Gian Sanjaya · 1 year, 11 months ago

Oh okay, thanks for the suggestion. I'll edit it.Log in to reply

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? – Calvin Lin Staff · 1 year, 11 months ago

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

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:

\(\displaystyle \lim_{x\rightarrow\infty} \Big(\sqrt[n]{P_a(x)} - \sqrt[m]{P_b(x)}\Big)=\frac{a_{n-1}}{nc^{n-1}}-\frac{b_{m-1}}{mc^{m-1}}\) – Gian Sanjaya · 1 year, 11 months ago

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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. – Calvin Lin Staff · 1 year, 11 months ago

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