Evaluation of the limit $\displaystyle\lim_{x\to\infty}x-\sqrt{(x-a)(x-b)}$

Question

I'm trying to evaluate the limit $\displaystyle\lim_{x\to\infty}x-\sqrt{(x-a)(x-b)}$

Since there is a square root of a product namely $\sqrt{(x-a)(x-b)}$ I think it's useful to use $\text{GM}\le\text{AM}$ for $(x-a)$ & $(x-b)$: $\sqrt{(x-a)(x-b)}\le\dfrac{(x-a)+(x-b)}{2}=x-\dfrac{a+b}{2}\tag*{}$

To see whether the function $f(x)=x-\sqrt{(x-a)(x-b)}$ is increasing or decreasing,

\begin{align}f'(x)&=1-\dfrac{2x-(a+b)}{2\sqrt{(x-a)(x-b)}}\\&=1-\dfrac{\dfrac{x-(a+b)}2}{\sqrt{(x-a)(x-b)}}\\&=1-\dfrac{\text{AM}}{GM}\\&\le0 \because\text{ GM}\le\text{AM}\end{align}

hence $f(x)=x-\sqrt{(x-a)(x-b)}$ is decreasing. Then in order for the existence of the limit, $f(x)$ must be bounded below.

Here I again use the same inequality:

\begin{align}\\&\sqrt{(x-a)(x-b)}\le\dfrac{(x-a)+(x-b)}{2}=x-\dfrac{a+b}{2}\\&\implies f(x)=x-\sqrt{(x-a)(x-b)}\ge\dfrac{a+b}{2}\end{align}

If I'm right, then the limit exists and $\displaystyle\lim_{x\to\infty}x-\sqrt{(x-a)(x-b)}=\dfrac{a+b}2$

Another query is that in the inequality $\sqrt{(x-a)(x-b)}\le\dfrac{(x-a)+(x-b)}{2}$ the equality holds iff $x-a=x-b$- how to solve it for $x$? I want to know where $f(x)$ takes its minimum.

Answer 1

You are correct, but the limit is easier to evaluate if you consider $$x-\sqrt{(x-a)(x-b)}=\frac{x^2-(x-a)(x-b)}{x+\sqrt{(x-a)(x-b)}}= \frac{(a+b)-ab/x}{1+\sqrt{(1-a/x)(1-b/x)}}$$ for $x> \max(a,b,0)$.

Answer 2

Alternatively, do the change of variable $x = 1/t$ $$\lim_{x\to+\infty}x - \sqrt{(x-a)(x-b)} = \lim_{t\to 0^+}\frac{1 - \sqrt{1 - (a + b)t + abt^2}}t$$ and apply Taylor: $$\sqrt{1 - (a + b)t + abt^2} = 1 - \frac{a + b}2 t + \cdots$$

Answer 3

Hint:

Evaluation of the limit $\displaystyle\lim_{x\to\infty}x-\sqrt{(x-a)(x-b)}$,

Rationalizing the numerator, $$\lim_{t\to0^+}\dfrac{1-\sqrt{1-(a+b)t+abt^2}}t=\lim_{t\to0^+}\dfrac{1-(1-(a+b)t+abt^2)}t\cdot\lim_{t\to0^+}\dfrac1{1+\sqrt{1-(a+b)t+abt^2}}$$

As $t\to0,t\ne0,\dfrac{1-(1-(a+b)t+abt^2)}t=(a+b)-abt$