Find the limit: $\displaystyle\lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x})$.

Question

Find the following limit: $$ \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )$$

I tried to simplify using conjugation. This gave me the following: $$ \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} $$

When I plug in the $\infty$, I'm left with $ \frac{\infty}{\infty} $. Did I mess up somewhere, or does the limit not exist?

Answer 1

You are almost there:

$$\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{x \sqrt{1+\frac{1}{x}}+x\sqrt{1-\frac{1}{x}}}$$

so...

Answer 2

Hint:

$$ax^2+bx+c\approx ax^2, ~~x\to\infty$$ so if $x\to+\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=x$ and if $x\to-\infty$ then $\sqrt{x^2+x}\approx\sqrt{x^2}=|x|=-x$

Answer 3

For $x\ge 1$ we have $$ \sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}=\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}. $$ It follows that $$ \lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}=1. $$