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How do I get from the first expression to the second?

The only reason the limits are included is because on WolframAlpha it mentioned that this was the case for large negative numbers of x:

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

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2 Answers 2

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Multiply the numerator and denominator by $-1$: $$\lim_{x\to\ -\infty} \frac{\sqrt{(x^2 + 2)}}{x} = \lim_{x\to\ -\infty} \frac{-\sqrt{(x^2 + 2)}}{-x}=\lim_{x\to\ -\infty} \frac{-\sqrt{(x^2 + 2)}}{|x|}=\\ =\lim_{x\to\ -\infty} \frac{-\sqrt{(x^2 + 2)}}{\sqrt{x^2}}=\lim_{x\to\ -\infty} -\sqrt{\frac{(x^2 + 2)}{x^2}}$$

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  • $\begingroup$ Thank you. Its super clear now. $\endgroup$
    – uhakdt
    Nov 17, 2018 at 1:08
  • $\begingroup$ You are welcome. Good luck. $\endgroup$
    – farruhota
    Nov 17, 2018 at 2:38
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Notice that we have $$\sqrt{x^2}=|x|$$

when $x<0$, we have $|x|=-x$

$$x=-|x|=-\sqrt{x^2}$$

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