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

The answer is 2, I don't know how to get it though. Thanks for your help.

Git Gud
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user3247608
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    Try multiplying the numerator and denominator by the conjugate. – Michael Burr Apr 01 '15 at 09:50
  • @user3247608, See http://math.stackexchange.com/questions/1205475/how-to-evaluate-this-limit-question-infinity/1205503#1205503 and http://math.stackexchange.com/questions/1125334/help-with-lim-x-rightarrow-infty-x2-sqrtx4-x2-1/1125342#1125342 – lab bhattacharjee Apr 01 '15 at 11:18

2 Answers2

2

$$\left(\sqrt{x^2+4x+1}-x\right)\cdot\frac{\sqrt{x^2+4x+1}+x}{\sqrt{x^2+4x+1}+x}=\frac{4x+1}{\sqrt{x^2+4x+1}+x}=$$

$$\frac{4x+1}{\sqrt{x^2+4x+1}+x}\cdot\frac{\frac1x}{\frac1x}=\frac{4+\frac1x}{\sqrt{1+\frac4x+\frac1{x^2}}+1}\xrightarrow[x\to\infty]{}...$$

Timbuc
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1

$(x^2+4x+1)^{1/2} = x(1+ 4/x+ 1/x^2)^{1/2}$

Using Taylor's theorem $$ (1+ 4/x+ 1/x^2)^{1/2} = 1 + 2/x + \mathcal{O}(x^{-2}) $$ So $$ (x^2+4x+1)^{1/2} = x + 2 + \mathcal{O}(x^{-1}) $$ The result is now clear.

Mark Joshi
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