Calculate $\lim_{x \to \infty} x - \sqrt{x^2 + 2x}$ without derivations

Question

How can I calculate $\displaystyle \lim_{x \to \infty} x - \sqrt{x^2 + 2x}$?

Here is what I've done so far:

Multiplying by $\displaystyle \frac{x + \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}}$

I got $\displaystyle \frac {-2x}{x+\sqrt{x^2 + 2x}}$

Multiplying this by $\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$

I got $\displaystyle \frac{-2}{1+\frac{\sqrt{x^2 + 2x}}{x}}$

I know I'm very close to the answer wich is $-1$, but I have no idea what to do next. I can´t just say that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1$, as far as I know...

Answer 1

Try: $$ \frac{\sqrt{x^2 + 2x}}{x} = \sqrt{\frac{x^2 + 2x}{x^2}} = \sqrt{1 + \frac{2}{x}} $$ By continuity of the square root function this means $$ \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1. $$

Answer 2

Do $\frac{\sqrt{x^2+2x}}{x}=\frac{x\sqrt{1+2/x}}{x}=\sqrt{1+2/x}$ now you can justify that the limit of this quotient is 1.

Answer 3

Note that $x - \sqrt{x^2 + 2x} = x - \sqrt{(x+1)^2 - 1}$. For large $x$, the second term is approximately $x+1$. That is to say, for $x > -1$, $$x - \sqrt{(x+1)^2 - 1} = x - (x+1)\sqrt{1 - \frac{1}{(x+1)^2}}$$ and the limit of the term in the square root is clearly $1$.

Answer 4

Another similar approach would be to write $$x - \sqrt{x^2 + 2x} =x- x \sqrt{1+\frac {2}{x}}$$ and to remember that, when $y$ is small compared to $1$, $\sqrt{1+y} \simeq 1+\frac{y}{2}$ (this is the start of the Taylor series). Now, replace $y$ by $\frac {2}{x}$, develop and simplify to get your limit of $-1$.