Find $\displaystyle \displaystyle \lim_{x \rightarrow - \infty} \left(\sqrt{x^2 + 2x} - \sqrt{x^2 + 1} \right)$.
$\displaystyle f(x) = \sqrt{(x+1)^2 -1} - \sqrt{x^2 +1}$
$\displaystyle f(x)= \frac{(x^2+2x) - (x^2+1)}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}} $
$\displaystyle f(x)= \frac{2x-1}{\sqrt{(x+1)^2 -1} + \sqrt{x^2 +1}} $
$\displaystyle f(x)= \frac{2-\frac{1}{x}}{\sqrt{(1+\frac{1}{x})^2 -\frac{1}{x^2}} + \sqrt{1+\frac{1}{x^2}}}}
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I saw in a book that it had been solved like this:
$\displaystyle \displaystyle\lim_{x \to -\infty} \sqrt{x^2+2x}-\sqrt{x^2+1}=\frac{x^2+2x-(x^2+1)}{\sqrt{x^2+2x}+\sqrt{x^2+1}}=\frac{2x-1}{\sqrt{x^2+2x}+\sqrt{x^2+1}} =$
$\displaystyle \frac{2x-1}{\sqrt{x^2(1+\frac{2}{x})}+ \sqrt{x^2(1+\frac{1}{x^2})}}= \displaystyle\lim_{x \to -\infty} \left \frac{2x-1}{|x|+|x|}= \frac{2x-1}{-2x}=-1 $
is it right?