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Currently I'm self studying limits. but I don't know how to get the answer to this question:
$$\lim _ { x\to \infty }\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)$$

can someone help me

Ramith Hettiarachchi
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    In general, whenever you see a limit of the form $\lim\sqrt{P(x)}-\sqrt{Q(x)}$, where $P(x)$ and $Q(x)$ are some polynomial functions, multiply it by the conjugate over the conjugate: $\tfrac{\sqrt{P(x)}+\sqrt{Q(x)}}{\sqrt{P(x)}+\sqrt{Q(x)}}$. Then either do a substitution, or divide the numerator and denominator by $x$ (or $x^2$... it depends on the context). – Hakim Sep 28 '14 at 10:30
  • yes. now i get it. – Ramith Hettiarachchi Sep 28 '14 at 10:42

3 Answers3

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So our limit is:

$$\lim _ { x\to \infty }\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)$$

We can rationalize the function by multiplying by the conjugate.

$$\lim _ { x\to \infty }\left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right) = \left(\sqrt{x^2+x+1}-\sqrt{x^2+1}\right)*\frac{\left(\sqrt{x^2+x+1}+\sqrt{x^2+1}\right)}{\left(\sqrt{x^2+x+1}+\sqrt{x^2+1}\right)}$$

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

Now we divide both numerator and denominator by $x$.

$$\lim _ { x\to \infty }\frac{x}{\left(\sqrt{x^2+x+1}+\sqrt{x^2+1}\right)} = \frac{1}{\left(\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}+\sqrt{\frac{x^2}{x^2}+\frac{1}{x^2}}\right)} = \frac{1}{\sqrt{1+0+0} + \sqrt{1+0}} = \frac{1}{2}$$

Varun Iyer
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Set $\dfrac1x=h,$

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

$$=\lim_{h\to0^+}\frac{\sqrt{1+h+h^2}-\sqrt{1+h^2}}h$$

Now rationalize the numerator

lab bhattacharjee
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Hint:

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

David H
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