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The problem is evaluate

$$\lim_{x\to\infty}(x^2-\sqrt{x^4 + 7x^2 + 1})$$

I understand all of the calculus involved, but am having trouble figuring out how to get started with the algebra. I have tried factoring and using conjugates, but the only answer I am able to get is $-7$, which is incorrect. Any help would be appreciated.

What I have done so far:

$$\frac{(x^2-\sqrt{x^4+7x^2+1})(x^2+\sqrt{x^4+7x^2+1})}{ x^2+\sqrt{x^4+7x^2+1}}$$

results in

$$\frac{-7x^2-1}{x^2+\sqrt{x^4+7x^2+1}}$$

factor out the $x^4$ under the radical, then divide numerator and denominator by $x^2$ to get

$$\frac{-7-1/x^2}{1 + \sqrt{1+7/x^2+1/x^4}}$$

at this point the limit as x approaches infinity would be -7/2 or -3.5.

  • 0
    Sounds good. The $\sqrt{x^4+..} > x^2$, so you may get negative answer. Include your calculation here2012-09-29
  • 2
    Do you forgot the $2$ in the denominator?2012-09-29
  • 0
    yes! thank you.2012-09-29
  • 0
    Replace $x^2$ by $x$ and use http://math.stackexchange.com/questions/30040/limits-how-to-evaluate-lim-limits-x-rightarrow-infty-sqrtnxna-n-1/30065#300652012-09-30

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