Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
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.

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

4 Answers 4

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.

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

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 $x \to \infty$.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.