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$$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}$$

I can solve it using l'Hôpital but just cannot find a way to do it algebraically.

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What limit are you trying to find? – Chris Eagle Feb 18 '13 at 21:19
Limit as $x$ goes to what? – Jim Feb 18 '13 at 21:19
@user62872 No limit, no question. It's an easy downvote. But I'll refrain from doing that to give you time to properly ask the question. – Git Gud Feb 18 '13 at 21:20
I am sorry limit as x approaches 0, couldnt find how to edit the original post – user62872 Feb 18 '13 at 21:20
@user62872 No problem. Don't forget to upvote answers which you find helpful and accept your favorite one. – Git Gud Feb 18 '13 at 21:24
up vote 3 down vote accepted

Multiply the top and bottom by $2\sqrt{x + 1} + x + 2$.

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$$\begin{align} \lim_{x\to 0}\frac{2\sqrt{x+1}-x-2}{x^2} &= \lim_{x\to 0}\frac{2\sqrt{x+1}-(x+2)}{x^2} \frac{2\sqrt{x+1}+(x+2)}{2\sqrt{x+1}+x+2}\\ &= \lim_{x\to 0}\frac{4(x+1)-(x+2)^2}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\ &= \lim_{x\to 0}\frac{4(x+1)-(x^2+4x+4)}{x^2} \frac{1}{2\sqrt{x+1}+x+2}\\ &= \lim_{x\to 0}\frac{-x^2}{x^2}\frac{1}{2\sqrt{x+1}+x+2}\\ &=\lim_{x\to 0}\frac{-1}{2\sqrt{x+1}+x+2}\\ &=-\frac{1}{4} \end{align}$$

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There's a typo in the last $\lim$. Nice answer. – Git Gud Feb 18 '13 at 21:35
I believe you left out a negative somewhere, nice work though – user62872 Feb 18 '13 at 21:36

$$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2} = \lim_{x\to 0} \dfrac{2\sqrt{x+1} - (x + 2)}{x^2}$$

To start, multiply numerator and denominator by the conjugate $$2\sqrt{x + 1} + (x + 2)$$

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Thank you, I had a dumb parentheses mistake – user62872 Feb 18 '13 at 21:37
You're welcome, user62872! – amWhy Feb 18 '13 at 21:53

We have $\sqrt{1+x}=1+\frac{1}{2}x+(\frac{1}{2})(\frac{-1}{2})\frac{x^2}{2}+o(x^2)$, so $$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}=\lim_{x \to 0}\dfrac{-\frac{1}{4}x^2+o(x^2)}{x^2}=-\frac{1}{4}$$

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