<s>[INST] Getting incorrect result on limit of square root I have a limit problem:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
According to Wolfram|Alpha the answer is $\frac{1}{2}$
However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:
$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \to \infty} \sqrt{x^2(1 + \frac{1}{x})} - x = \lim_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x = \lim_{x \to \infty} x(\sqrt{1 + \frac{1}{x}} - 1)$
Let $t = \frac{1}{x}$
$\lim_{x \to 0} \frac{\sqrt{1 + t} - 1}{t} \Rightarrow (l'Hopital) \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + t}}}{1} = \lim_{x \to 0} \frac{1}{\sqrt{1+t}} = 1$
 [/INST] You've lost your 2 in the derivation!
$\frac{d}{dt}\sqrt{1+t} = \frac{1}{2}\frac{1}{\sqrt{1+t}}$
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