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 $t = \frac{1}{x}$
$\lim_{x \to 0} \frac{\sqrt{1 + t} - 1}{t} \Rightarrow (l'Hopital) \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + t}}}{1} = \lim_{x \to 0} \frac{1}{\sqrt{1+t}} = 1$

$\lim_{x \to \infty} \sqrt{x^2 + x} - x$
approach:
taking $t = \frac{1}{x}$ hence t tends to 0
So,
$\lim_{t \to 0} \sqrt{\frac{1}{t^2} + \frac{1}{t}} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{\sqrt{1+t}}{t} - \frac{1}{t}$
$= \lim_{t \to 0} \frac{1}{t}  (\sqrt{1+t} - 1)$
multiplying by $\frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1}$
$= \lim_{t \to 0} \frac{1}{\sqrt{1+t} + 1}$
Now apply limits
