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?
 A: Hint:
$$\sqrt{x^2+4x}- x = \frac{4x}{\sqrt{x^2+4x}+x}$$
And you are right with the answer of $2$.
A: 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 $2$.
A: Just to give an alternative to multiplying by the conjugate expression, note that letting $x=u-2$ gives
$$\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=2+\lim_{u\to\infty}(\sqrt{u^2-4}-u)$$
Now for $u\gt0$,
$$\left(u-{2\over u}\right)^2=u^2-4+{4\over u^2}\lt u^2-4\implies u-{2\over u}\lt\sqrt{u^2-4}\lt u$$
hence
$$-{2\over u}\lt\sqrt{u^2-4}-u\lt0$$
so by the Squeeze Theorem $\lim_{u\to\infty}(\sqrt{u^2-4}-u)=0$, and thus $\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=2$.
Added later:  Here is another alternative.  Eschewing motivation, letting $x={1\over u+u^2}$ with $u\gt0$ gives
$$\sqrt{x^2+4x}-x=\sqrt{\left(1\over u+u^2\right)^2+{4\over u+u^2}}-{1\over u+u^2}={\sqrt{1+4u+4u^2}-1\over u+u^2}={\sqrt{(1+2u)^2}-1\over u+u^2}={1+2u-1\over u+u^2}={2\over1+u}$$
and thus
$$\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=\lim_{u\to0^+}{2\over1+u}=2$$
(To motivate things, first let $x=1/h$ to obtain $(\sqrt{1+4h}-1)/h$ and then let $h=u+u^2$ to get a square inside the square root.)
A: Hint
$$ \lim_{x\rightarrow \infty}(\sqrt{x^2+4x}-x)\frac{\sqrt{x^2+4x}+x}{\sqrt{x^2+4x}+x}=
\lim_{x\rightarrow \infty}\frac{x^2+4x-x^2}{\sqrt{x^2+4x}+x}=
\lim_{x\rightarrow \infty}\frac{4x}{x\sqrt{1+\frac{4}{x}}+x}=...$$
A: Putting $\displaystyle\frac1x=h,$
$$\lim_{x\to\infty}\sqrt{x^2+4x} - x=\lim_{h\to0^+}\left(\sqrt{\frac{1+4h}{h^2}}-\frac1h\right)=\lim_{h\to0^+}\left(\frac{\sqrt{1+4h}-1}h\right)$$
$$=\lim_{h\to0^+}\frac{1+4h-1}{h(\sqrt{1+4h}+1)}=4\lim_{h\to0^+}\frac1{\sqrt{1+4h}+1}\text{ as }h\ne0\text{ as }h\to0^+$$
$$=\cdots$$
