Expressing a divergent function in algebraic terms

Question

We're asked to find the $\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x })\cos(x)$. I've tried to break down the expression a little bit but couldn't get far. The graph clearly shows that the function oscillates and thus the limit doesn't exist but how would one get to the conclusion algebraically? Perhaps we could do so with the Squeeze theorem by stating that $\cos(x)\leq1$

My attempt:

$$=\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x })\cos(x)$$ $$=\lim_{x\to\infty}x\sqrt{\Big(1+\frac{1}{x}\Big)}-x\sqrt{\Big(1-\frac{1}{x}\Big)}\cos(x)$$

Answer 1

Use the fact that\begin{align}\sqrt{x^2+x}-\sqrt{x^2-x}&=\frac{\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}{\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}}\end{align}

Answer 2

By binomial first order expansion we have that

• $\sqrt{x^2+x}=x(1+1/x)^\frac12=x+\frac12+o(1)$
• $\sqrt{x^2-x}=x(1-1/x)^\frac12=x-\frac12+o(1)$

and then

$$\sqrt{x^2+x}-\sqrt{x^2-x }=x+\frac12-x+\frac12+o(1)=1+o(1) \to 1$$

then consider

• $x_n = 2\pi n \to \infty$

$$(\sqrt{x_n^2+x_n}-\sqrt{x_n^2-x_n})\cos(x_n) \to 1\cdot 1=1$$

but for

• $x_n = \frac{\pi}2+2\pi n \to \infty$

$$(\sqrt{x_n^2+x_n}-\sqrt{x_n^2-x_n})\cos(x_n) \to 1\cdot 0=0$$

therefore, since there exist two subsequnces with different limit, the given limit doesn't exist.