How to prove that $\lim_{x\rightarrow\infty}x-\sqrt{x(x+1)}=-\frac12$

Question

How do I prove that $$\lim_{x\rightarrow\infty}x-\sqrt{x(x+1)}=-\frac12$$? I first got this question when I was graphing $\sqrt{x(x+1)}$ and I noticed that it approached a linear function as $x$ increases. We couldn't use L'hopital's rule since these functions aren't in fraction form. Also, we couldn't simply substitute $x=\infty$. And Taylor series approximation seems very complicated.

Answer 1

\begin{align*} &\lim\limits_{x\rightarrow \infty} x-\sqrt{x(x+1)} \\ =& \lim\limits_{x\rightarrow \infty} \frac{x^2-x(x+1)}{x+\sqrt{x(x+1)}} \\ =& \lim\limits_{x\rightarrow \infty} - \frac{x}{x+\sqrt{x(x+1)}} \\ =& \lim\limits_{x\rightarrow \infty} - \frac{1}{1+\sqrt{1+\frac1x}} = -\frac12 \end{align*}

Usually, limits of the form $\infty-\infty$ can be calculated by multiplying the numerator and denominator by something of the form $\infty+\infty$ and then simplifying.

Hope this helps you.

Answer 2

You can multiply and divide that difference by its conjugate to get a difference of squares in the numerator, end up with a fraction with numerator $x$, and then divide by $x$ every term to end up with a limit you know how to operate.

Answer 3

$x - \sqrt{x(x+1)}$

$ = x(1 - \sqrt{ {x(x+1) \over x^2} })$

$ = x(1 - \sqrt{ 1 + {1 \over x}})$

$ = x(1 - \left[ 1 + {1 \over 2x} - {1 \over 8x^2} + \ldots \right])$

$ = x(-{1 \over 2x} + {1 \over 8x^2} + \ldots )$

$ = -{1 \over 2} + {1 \over 8x} + ...$

When you take $\lim_{x \rightarrow \infty}$, the first term $-{1 \over 2}$ survives and other terms $\rightarrow 0$.

Answer 4

For the term $ \ \sqrt{x(x+1)} \ \ , \ $ one can also "complete the square" under the radical to obtain $ \ \sqrt{x^2 + x } $ $ = \ \sqrt{\left( x + \frac12 \right)^2 \ - \ \frac14} \ \ . \ $ The curve for this function is then the "upper half" of the hyperbola $ \ \left(x + \frac12 \right)^2 \ - \ y^2 \ = \ \frac14 \ \ , \ $ for which the asymptotes are $ \ y \ = \ | \ x + \frac12 \ | \ \ . \ $ The "limits at infinity" for our function are then $$ \lim_{x \ \rightarrow \ +\infty} \ x \ - \ \sqrt{x(x+1)} \ \ \rightarrow \ \ \lim_{x \ \rightarrow \ +\infty} \ x \ - \ \left| \ x + \frac12 \ \right| \ \ = \ \ \lim_{x \ \rightarrow \ +\infty} \ x \ - \ \left( \ x + \frac12 \ \right) \ \ = \ \ -\frac12 \ \ , $$ $$ \lim_{x \ \rightarrow \ -\infty} \ x \ - \ \sqrt{x(x+1)} \ \ \rightarrow \ \ \lim_{x \ \rightarrow \ -\infty} \ x \ + \ \left( \ x + \frac12 \ \right) \ \ = \ \ \lim_{x \ \rightarrow \ -\infty} \ 2x \ + \ \frac12 \ \ \ = \ \ -\infty \ \ . $$

Answer 5

Let $y=\sqrt{x^2+x}$, then replace $y\to\dfrac1y$ to recover a derivative. $$\begin{align*} \lim_{x\to\infty} \left(x-\sqrt{x^2+x}\right) &= \lim_{y\to\infty}\left(\frac{\sqrt{4y^2+1}-1}2-y\right) \\ &= -\frac12 + \lim_{y\to\infty} y \left(\sqrt{1+\frac1{4y^2}} - 1\right) \\ &= -\frac12 + \lim_{y\to0^+} \frac{\sqrt{1+\frac{y^2}4}-1}{y} \\ &= -\frac12 + \frac{d\sqrt{1+\frac{y^2}4}}{dy}\bigg|_{y=0} = \boxed{-\frac12} \end{align*}$$