Limit at infinity of two root functions

Question

What is the limit of $f(x) = \sqrt{|x|} - \sqrt{|x-1|}$ as $x\to\infty$? I can plot this and see that $f(x)$ is going to zero but I would like to solve this algebraically. I feel like I need to put both terms under the same root.

Answer 1

Since, as $x \to \infty$, $x$ eventually becomes greater than $0$, we can drop the absolute values without a problem.

Consider multiplying and dividing by $\sqrt{x} + \sqrt{x-1}$. Then on the top you can utilize the difference of squares formula, i.e. $(a-b)(a+b) = a^2 - b^2$, and see that

$$\begin{align} \lim_{x \to \infty} \sqrt{x} - \sqrt{x-1} &= \lim_{x \to \infty} (\sqrt{x}- \sqrt{x-1}) \cdot \frac{\sqrt{x}+ \sqrt{x-1}}{\sqrt{x}+ \sqrt{x-1}} \\ &= \lim_{x \to \infty}\frac{1}{\sqrt{x}+\sqrt{x-1}} \end{align}$$

The limit of this final expression is clearly $0$, and thus

$$\lim_{x \to \infty} \sqrt{x} - \sqrt{x-1} = 0$$

Answer 2

$f(x)=\sqrt{|x|}-\sqrt{|x-1|}=\frac{(\sqrt{|x|}-\sqrt{|x-1|})(\sqrt{|x|}+\sqrt{|x-1|})}{\sqrt{|x|}+\sqrt{|x-1|}}=\frac{|x|-|x-1|}{\sqrt{|x|}+\sqrt{|x-1|}}$

so $\frac{-1}{\sqrt{|x|}+\sqrt{|x-1|}}\le f(x)\le \frac{1}{\sqrt{|x|}+\sqrt{|x-1|}}$

It goes to zero as x goes to $\infty$

Answer 3

Using the binomial series, $\lim\limits_{x\to\infty} \sqrt x-\sqrt {x-1}=\lim\limits_{x\to\infty}\sqrt x \left(1-\sqrt{1-\dfrac 1x}\right)$

$=\lim\limits_{x\to\infty}\sqrt x\left(1-\left(1-\dfrac12\dfrac1x+O\left(\dfrac1{x^2}\right)\right)\right)=\lim\limits_{x\to\infty}\sqrt x\left( \dfrac12O\left(\dfrac1x\right)\right)=0.$