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

Question

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?

Answer 1

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 $2$.

Answer 2

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

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

Answer 3

Hint

$$ \lim_{x\rightarrow \infty}(\sqrt{x^2+4x}-x)\frac{\sqrt{x^2+4x}+x}{\sqrt{x^2+4x}+x}= \lim_{x\rightarrow \infty}\frac{x^2+4x-x^2}{\sqrt{x^2+4x}+x}= \lim_{x\rightarrow \infty}\frac{4x}{x\sqrt{1+\frac{4}{x}}+x}=...$$

Answer 4

Putting $\displaystyle\frac1x=h,$

$$\lim_{x\to\infty}\sqrt{x^2+4x} - x=\lim_{h\to0^+}\left(\sqrt{\frac{1+4h}{h^2}}-\frac1h\right)=\lim_{h\to0^+}\left(\frac{\sqrt{1+4h}-1}h\right)$$

$$=\lim_{h\to0^+}\frac{1+4h-1}{h(\sqrt{1+4h}+1)}=4\lim_{h\to0^+}\frac1{\sqrt{1+4h}+1}\text{ as }h\ne0\text{ as }h\to0^+$$

$$=\cdots$$