# Algebraically Solve Limit

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

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

• What limit are you trying to find? Feb 18, 2013 at 21:19
• Limit as $x$ goes to what?
– Jim
Feb 18, 2013 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. Feb 18, 2013 at 21:20
• I am sorry limit as x approaches 0, couldnt find how to edit the original post Feb 18, 2013 at 21:20
• @user62872 No problem. Don't forget to upvote answers which you find helpful and accept your favorite one. Feb 18, 2013 at 21:24

\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}

• There's a typo in the last $\lim$. Nice answer. Feb 18, 2013 at 21:35
• I believe you left out a negative somewhere, nice work though Feb 18, 2013 at 21:36

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

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

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

• Thank you, I had a dumb parentheses mistake Feb 18, 2013 at 21:37
• You're welcome, user62872! Feb 18, 2013 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 $$\lim_{x \to 0} \dfrac{2\sqrt{x+1}-x-2}{x^2}=\lim_{x \to 0}\dfrac{-\frac{1}{4}x^2+o(x^2)}{x^2}=-\frac{1}{4}$$