Difficult limit evaluation: $\lim_{x\to\infty}(\sqrt{x^2+4x} - x)$ I'm struggling to find the solution of the following:
$$\lim_{x\rightarrow\infty}(\sqrt{x^2+4x} - x)$$
I come to the answer of $0$.
The book has an answer of $4/1$.
The book explains a part of the question briefly. Looking at the brief answer information given I then come to an answer of $2$...
What is the answer?

Multiplying by the quantity $\sqrt{x^2 + 4x} + x$ on top and bottom leads to
$$\sqrt{x^2 + 4x} - x = \frac{x^2 + 4x - x^2}{\sqrt{x^2 + 4x} + x} = \frac{4x}{\sqrt{x^2 + 4x} + x}$$
Now divide each term by $x$, noting that this becomes $x^2$ under the square root:
$$\frac{4x}{\sqrt{x^2 + 4x} + x} = \frac{\frac{4x}{x}}{\sqrt{\frac{x^2 + 4x}{x^2}} + \frac{x}{x}} = \frac{4}{1 + \sqrt{1 + \frac{4}{x}}}$$
Taking $x \to \infty$, this limit is $2$. There appears to be an error in the book, since Wolfram Alpha also returns $2$.
