Difficult limit evaluation: $\lim_{x\to\infty}(\sqrt{x^2+4x} - x)$

Hint: $$\sqrt{x^2+4x}- x = \frac{4x}{\sqrt{x^2+4x}+x}$$

And you are right with the answer of $2$.

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 $2$.