# How do I find this limit: $\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$ [duplicate]

$$\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$$

The answer is $$\frac{-3}{2}$$ according to Wolfram alpha.

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## marked as duplicate by Aryabhata, Micah, Amzoti, Davide Giraudo, vonbrandApr 12 '13 at 16:09

*find this limit – Jp McCarthy Apr 12 '13 at 14:59

I would consider $$\lim_{x \to \infty} \sqrt{x^4-3x^2-1}-x^2$$ as $$\lim_{x \to \infty} \dfrac{\sqrt{x^4-3x^2-1}-x^2}{1}$$

Then multiply $\sqrt{x^4-3x^2-1}+x^2$ top and bottom to get rid of the ridiculous looking square root sign on the top. Then bingo!

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Putting $x=\frac1h$ as $x\to\infty, h\to0$

$$\lim_{x \to \infty} \left(\sqrt{x^4-3x^2-1}-x^2 \right)$$

$$=\lim_{h\to0}\left(\frac{\sqrt{1-3h^2-h^4}-1}{h^2}\right)$$

$$=\lim_{h\to0}\left(\frac{\{1-h^2(h^2+3)\}^\frac12-1}{h^2}\right)$$

$$=\lim_{h\to0}\left(\frac{1-\frac12h^2(h^2+3)+O(h^4)-1}{h^2}\right)$$

$$=\lim_{h\to0} \left(-\frac12(h^2+3)+O(h^2)\right)$$

$$=-\frac32$$

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first, substitute : $t=x^2$, you get :

$$\lim_{t\to +\infty} \sqrt{t^2-3t-1}-t=\lim_{t\to \infty} \frac{-3t-1}{\sqrt{t^2-3t-1}+t}=\lim_{t\to \infty} \frac{-3-\frac{1}{t}}{\sqrt{1-\frac{3}{t} -\frac{1}{t^3}}+1} =\frac{-3}{2}.$$

Explanation : first we use this identity for $a\neq -b$ : $a-b =\frac{a^2-b^2}{a+b}$, then we factor $t$ from the numerator and the denomenator, the limit of the numerator is $-3$ and the limit of the numerator is $2$.

Further more, for any reals $a,b$ : $$\lim_{x\to+ \infty } \sqrt{x^2+ax+b}-x=\frac{a}{2}.$$

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Hint: Can you relate the expression under the root sign to $x^2$ in some way? That might make it easier to see where the value goes as $x\to\infty$.

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