<s>[INST] Dividing the function by the largest power of x to rationalize(infinity minus infinity) 
Calculating the limits of the above when x approaches infinity, why can't we just divide the function by the largest power of x and then say inside of the square root should be equal to "1" when x is at infinity level? 

This seems so right to me but I know it is dead wrong. Could somebody tell me what I am missing? 
 [/INST] Since $\lim_{x\to \infty}|x|\sqrt{1\pm\frac 2x}\to \infty$, we have $+\infty-\infty$. We cannot find the limit in this form.
Multiplying
$$\sqrt{x^2+2x}-\sqrt{x^2-2x}$$
by
$$\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\ (=1)$$
gives you
$$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-2x})(\sqrt{x^2+2x}+\sqrt{x^2-2x})}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$
$$=\lim_{x\to \infty}\frac{(x^2+2x)-(x^2-2x)}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}=\lim_{x\to \infty}\frac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$$$=\lim_{x\to\infty}\frac{\frac{4x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-2x}}{x}}=\lim_{x\to\infty}\frac{4}{\sqrt{\frac{x^2+2x}{x^2}}+\sqrt{\frac{x^2-2x}{x^2}}}$$$$=\lim_{x\to \infty}\frac{4}{\sqrt{1+\frac 2x}+\sqrt{1-\frac 2x}}=\frac{4}{\sqrt 1+\sqrt 1}=2.$$
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