$$ \lim_{x \to \infty} \left(\sqrt{x^2+x+1}-\sqrt{x^2-1} \right) $$
I have tried rationalizing but that just introduces a extra denominator. I have tried squaring and rooting but that give me $\lim_{ x \to \infty} \sqrt{x}$ which is not the answer.
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That denominator creates a $\infty/\infty$ form, which is easy to evaluate. – Dec 02 '17 at 10:57
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1See https://math.stackexchange.com/questions/2056840/solving-lim-x-to-infty-sqrtx2x1-sqrt1-xx2-using-o-littl – lab bhattacharjee Dec 02 '17 at 11:13
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You can use binomial approximation $$(1+a)^b \approx 1+ba$$ for $a \to 0$.
This will give you (at $x \to \infty$)$$ \sqrt{x^2+x+1} = x\left(\sqrt{ 1+\underbrace{\frac 1x+ \frac1{x^2}}_{\to 0}}\right) \approx x\left(1+\frac1{2x}+\frac{1}{2x^2} \right)$$
And $$\sqrt{x^2-1}=x\left( \sqrt{1\underbrace{-\frac{1}{x^2}}_{\to 0}}\right) \approx x\left(1-\frac1{2x^2}\right)$$
Thus $$\lim_{x \to \infty} \left(\sqrt{x^2+x+1}-\sqrt{x^2-1} \right)= \lim_{x \to \infty} x\left(1+\frac1{2x}+\frac{1}{2x^2} \right)- x\left(1-\frac1{2x^2}\right)=\frac 12$$
Jaideep Khare
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1I sincerely hate those who don't give justification of their down vote and more importantly on a post which is good enough to be upvoted. +1 from me. – Vidyanshu Mishra Dec 02 '17 at 14:58
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As you said, we first proceed by rationalising:
$$\sqrt{x^2+x+1}-\sqrt{x^2 - 1} = \frac{x+2}{\sqrt{x^2+x+1}+\sqrt{x^2 - 1} }$$
Now divide by $x$ and apply limit $x \to \infty$.
$$ \lim_{x\to \infty}\frac{1+\frac{2}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{1 - \frac{1}{x^2}} }$$
You will see the limit is $1/2$
jonsno
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Not only $\lim_{x\to\infty}\sqrt x$ is not the answer, but it's not even an answer: a limit when $x$ tends to some value is a number, and cannot depend on $x$.
You can indeed rationalise, and use some equivalents for the denominator: $$\sqrt{x^2+x+1}-\sqrt{x^2-1}=\frac{x^2+x+1-(x^2-1)}{\sqrt{x^2+x+1}+\sqrt{x^2-1}}=\frac{x+2}{\sqrt{x^2+x+1}+\sqrt{x^2-1}}$$ Now $x+2\sim_\infty x$ and $\sqrt{x^2+x+1},\sqrt{x^2-1}\sim_\infty |x|$, so $$\sqrt{x^2+x+1}-\sqrt{x^2-1}\sim_\infty \frac x{2|x|}\to\begin{cases}\phantom{-}\dfrac12&\text{if }x\to+\infty,\\[1ex]- \dfrac12&\text{if }x\to -\infty.\end{cases}$$
Bernard
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$$\lim_{x \to \infty} \left(\sqrt{x^2+x+1}-\sqrt{x^2-1} \right) = \lim_{x \to \infty} \frac{\left(\sqrt{x^2+x+1}-\sqrt{x^2-1} \right)\left(\sqrt{x^2+x+1}+\sqrt{x^2-1}\right)}{ \sqrt{x^2+x+1}+\sqrt{x^2-1} } $$
$$= \lim_{x \to \infty} \frac{x² + x +1 - x² +1}{ \sqrt{x^2+x+1}+\sqrt{x^2-1}}$$
$$= \lim_{x \to \infty} \frac{ x +2}{ \sqrt{x^2+x+1}+\sqrt{x^2-1}}$$
$$= \lim_{x \to \infty} \frac{ x +2}{ \sqrt{x²(1+1/x+1/x²)}+\sqrt{x^2(1-1/x²)}}$$
$$= \lim_{x \to \infty} \frac{ x(1 +2/x)}{ x\sqrt{1+1/x+1/x²}+x\sqrt{1-1/x²}}$$
$$= \lim_{x \to \infty} \frac{ 1+2/x}{ \sqrt{1+1/x+1/x²}+\sqrt{1-1/x²}} = \frac{1}{1+1} = \frac{1}{2}$$