Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

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What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$

In other words, if I am given a polynomial $P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0$, how would I find $$\lim_{x\rightarrow\infty} P(x)^{1/n}-x.$$

For example, how would I evaluate limits such as $$\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x$$ or $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$

Answer

Your limit can be rewritten as
$$\lim_{x\rightarrow\infty}\left(\frac{\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1}{1 \over x}\right)$$
Or equivalently,
$$\lim_{y\rightarrow 0}\left(\frac{\sqrt[n]{1+{a_{n-1}}{y}+\cdots+{a_{0}}{y^{n}}}-1}{y}\right)$$
This, by the definition of derivative, is the derivative of the function $f(y) = {\sqrt[n]{1+{a_{n-1}}{y}+\cdots+{a_{0}}{y^{n}}}}$ at $y = 0$, which evaluates via the chain rule to ${a_{n-1} \over n}$.

Attribution
Source : Link , Question Author : Eric Naslund , Answer Author : Zarrax

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