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

Consider the limit $$\lim_{x \to \infty} \sqrt{1+4x+x^{2}}-x$$ By completing the square inside the square root, we get $1+4x+x^2=(x+2)^{2}-3$. Thus our limit becomes $$\lim_{x \to \infty} \sqrt{(x+2)^2-3}-x=\lim_{x \to \infty} \sqrt{(x+2)^2}-x=\lim_{x \to \infty} |x+2|-x=2$$ I want to check if this argument is airtight. First, we say that as $x$ tends to infinity the constant term is negligible, then that since $x$ is obviously positive $|x+2|=x+2$. Is this idea of the constant term being negligible, and ignoring it in the limit, rigorous? It clearly works since the limit is indeed 2, but I wanted to know if this is a correct way to arrive at it.

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marked as duplicate by JavaMan, tomasz, Rudy the Reindeer, Nate Eldredge, Norbert Oct 6 '12 at 22:04

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up vote 9 down vote accepted

Well, I would multiply top and (missing) bottom by $\sqrt{1+4x+x^2}+x$.

And then we don't have the somewhat hand-wavy throwing away the $-3$ as negligible.

It is indeed negligible, but developing accurate intuition about what is negligible and what is not takes experience, and at this stage you would, in an analysis course, be expected to provide details. After all, one could also argue that $4x$ is negligible. It is tiny compared to $x^2$. But throwing it away gives the wrong answer.

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$$\lim_{x \to \infty} \sqrt{1+4x+x^{2}}-x=\lim_{x \to \infty} (\sqrt{1+4x+x^{2}}-x)\frac{\sqrt{1+4x+x^{2}}+x} { \sqrt{1+4x+x^{2}}+x}=$$ $$=\lim_{x \to \infty}\frac {{1+4x+x^{2}}-x^2}{\sqrt{1+4x+x^{2}}+x}=\lim_{x \to \infty}\frac {1+4x}{\sqrt{1+4x+x^2}+x}=$$ $$ =\lim_{x \to \infty}\frac {\frac{1+4x}{x}}{\frac {\sqrt{1+4x+x^{2}}+x} {x}}=\lim_{x \to \infty}\frac {\frac{1}{x}+4} {\sqrt{\frac{1}{x^2}+\frac{4}{x}+1} +1}=\frac{0+4}{\sqrt{0+0+1}+1}=\frac{4}{2}=2 $$

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Your argument is leaky, I'm afraid. It is true that if $\lim_{x \to \infty} f(x)$ and $\lim_{x \to \infty} g(x)$ both exist (and are finite), then $\lim_{x \to \infty} (f(x) - g(x))$ exists, and is equal to $\lim_{x \to \infty} f(x) - \lim_{x \to \infty} g(x)$. But this is not the case here: $\lim_{x \to \infty} \sqrt{(x+2)^2-3}$ is infinite.

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I don't really know about the step with the constant being negligible, of course it seems intuitive, but I couldn't prove it right away. Instead, try :

$|\sqrt{(x+2)^2-3}-x-2| = |\frac{-3}{\sqrt{(x+2)^2-3}+x+2}| \rightarrow 0 $ for $x \to \infty$. This seems better, as you can easily show that $\sqrt{(x+2)^2-3}+x+2 \to \infty$ for $x \to \infty$.

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