Intereting Posts

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

Here is what I’ve done so far:

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

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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…

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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.

$$

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.

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 $1$.

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 $-1$.

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