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The problem is: $$ \lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{x} $$ So far, I've tried substituting $\sqrt{1+x+x^2}-1$ with some variable $t$, but when $x\to0$, $t\to\sqrt{1}$.
I have also tried to rationalize the numerator, and applied l'hospital.
I simply can't figure out this limit.
Any help would be greatly appreciated! Thanks in advance.

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    $\begingroup$ also tried to rationalize the numerator And what happened there? $\endgroup$
    – dxiv
    Mar 7, 2017 at 20:13

4 Answers 4

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Rationalizing the numerator is the way to go. We have

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

Can you finish now?

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  • $\begingroup$ Yes, thank you for you help! I now realise I was making a trivial mistake. $\endgroup$ Mar 13, 2017 at 17:12
  • $\begingroup$ You're welcome! My pleasure. -Mark $\endgroup$
    – Mark Viola
    Mar 13, 2017 at 17:16
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$$\lim _{ x\to 0 } \frac { \sqrt { 1+x+x^{ 2 } } -1 }{ x } \cdot \frac { \sqrt { 1+x+x^{ 2 } } +1 }{ \sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { x+{ x }^{ 2 } }{ x\sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { 1+x }{ \sqrt { 1+x+x^{ 2 } } +1 } =\frac { 1 }{ 2 } $$

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Also you can use L'Hopital's rule: $$ \lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{x}=\frac00=\lim_{x\to0}\frac{\frac{d}{dx}\left[\sqrt{1+x+x^2}-1\right]}{\frac{d}{dx}\left[x\right]}=\lim_{x\to0}\frac{1+2x}{2\sqrt{1+x+x^2}}=\color{red}{\frac12} $$

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  • $\begingroup$ Thanks, this looks like the shortest method. $\endgroup$ Mar 13, 2017 at 17:14
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The binomial expansion for a square root is simple:

$$\sqrt{a+b}=(a+b)^{1/2}=a^{1/2}+\frac12b\cdot a^{-1/2}+\mathcal O(b^2\cdot a^{-3/2})$$

Thus,

$$\sqrt{1+x+x^2}=1^{1/2}+\frac12(x+x^2)1^{-1/2}+\mathcal O(x^2)$$

$$\begin{align}\frac{\sqrt{1+x+x^2}-1}x&=\frac{1+\frac12(x+x^2)-1+\mathcal O(x^2)}x\\&=\frac{\frac12(x+x^2)+\mathcal O(x^2)}x\\&=\frac12(1+x)+\mathcal O(x)\\&\to\frac12(1+0)=\frac12\end{align}$$

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  • $\begingroup$ Thank you for outlining a new method. This will come in handy. $\endgroup$ Mar 13, 2017 at 17:13
  • $\begingroup$ Yes. I find this quite useful, and furthermore simple enough to do in your head quite quickly :-) $\endgroup$ Mar 13, 2017 at 17:57

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