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When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get

$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac{1}{-x}}$ to get an undefined answer or $\frac{-7}{2}.$

My teacher's solution involves multiplying by $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$

$$=\lim_{x \to -∞}\frac{-7}{\sqrt{x^2/x^2+7x/x^2}+1}$$

$$=-\frac{7}{\sqrt{1+0}+1}$$

$$=\frac{-7}{2}$$

However, I multiplied

by $\frac{\frac{1}{x}}{\frac{1}{x}}$ and got the following:

$$\lim_{x \to -∞}\frac{7}{\sqrt{x^2/x^2+7x/x^2}-1}$$

$$\frac{7}{\sqrt{1+0}-1}$$

$$\frac{7}{0}$$

Which is undefined.

Why does multiplying by what is essentially $1$ cause different answers in general, and in particular for evaluating limits?

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    $\begingroup$ $-1\neq \sqrt{(-1)^2}$. The square root returns the principle square root, which in the case of non-negative reals is always non-negative. $\endgroup$ – JMoravitz Jul 27 '16 at 18:00
  • $\begingroup$ @JMoravitz it's not. I want to know specifically about its application to limits in this case, and which is correct. $\endgroup$ – Max Li Jul 27 '16 at 18:03
  • $\begingroup$ The application to limits is "don't make an invalid simplification". ​ ​ $\endgroup$ – user57159 Jul 27 '16 at 18:04
  • $\begingroup$ @RickyDemer Would you mind writing an answer saying which answer is correct? $\endgroup$ – Max Li Jul 27 '16 at 18:05
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    $\begingroup$ The answers which appeared below seem to do a good job of clearing things up. Here, it became even more muddied by the fact that we were approaching negative infinite as a limit (as opposed to positive infinite). The end result is that for reals, $\sqrt{x^2}=|x| = \begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$ $\endgroup$ – JMoravitz Jul 27 '16 at 18:32
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Your teacher has left out some justification, which is probably why you are confused.

In general, we have the following identity for all real $x$:

$$\sqrt{x^2}=|x|.\tag{1}$$ Also, for nonnegative real $a$ and positive real $b,$ we have $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac ab}.\tag{2}$$ Putting $(1)$ and $(2)$ together, if $x$ a non-zero real number and $a$ is a nonnegative real number, then $$\frac1{|x|}\sqrt{a}=\sqrt{\frac a{x^2}}.\tag{$\star$}$$


Your teacher took advantage of $(\star),$ without being explicit about it. In particular, since we're taking the limit as $x\to-\infty,$ then we may as well assume that $x\le-7$ (since it has to be, eventually), so that $x^2+7x$ is nonnegative, $x$ is non-zero, and $-x=|x|.$ Thus, we have $$\frac{1}{-x}\sqrt{x^2+7x}=\frac1{|x|}\sqrt{x^2+7x}=\sqrt{\frac{x^2+7x}{x^2}}.$$

On the other hand, $$\frac{1}{x}\sqrt{x^2+7x}=\frac1{-|x|}\sqrt{x^2+7x}=-\frac1{|x|}\sqrt{x^2+7x}=-\sqrt{\frac{x^2+7x}{x^2}}.$$ From there, you should come up with the same answer.

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  • $\begingroup$ Wow, that's a lot of tricky business to watch out for. Thanks. $\endgroup$ – Max Li Jul 27 '16 at 18:40
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    $\begingroup$ You're very welcome. Also, see related approaches here, here, here, and here. $\endgroup$ – Cameron Buie Jul 27 '16 at 18:40
  • $\begingroup$ You are my hero $\endgroup$ – Max Li Jul 27 '16 at 18:46
  • $\begingroup$ Well, we have (2) even when $a$ is non-negative real and $b$ is zero, since either both sides are undefined or both sides are unsigned infinity. ​ ​ $\endgroup$ – user57159 Jul 28 '16 at 10:46
  • $\begingroup$ @Ricky: Fair enough. $\endgroup$ – Cameron Buie Jul 28 '16 at 11:43
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Let's see what happens to $$f(x) = \sqrt{x^2 + 7x} + x$$ when $x$ takes on actual negative values. When $x = -100$, we get $$f(-100) = 10 \sqrt{93} - 100 \approx -3.56349.$$ When $x = -10000$, we get $$f(-10000) = 100 \sqrt{9993} - 10000 \approx -3.50061.$$ So, from a numerical standpoint, this seems to suggest that the limit should exist. When we rationalize the numerator, i.e. $$f(x) = \frac{7x}{\sqrt{x^2 + 7x} - x},$$ we are still okay, but the next step, division by $x$, requires care because when we write $$\frac{7}{\frac{1}{x}\sqrt{x^2 + 7x} - 1} = \frac{7}{\sqrt{1 + 7/x^2} - 1},$$ we inadvertently commit a sign error: this is because if $x < 0$, we no longer have $$\frac{1}{x} = \sqrt{\frac{1}{x^2}}.$$ The LHS is negative; the RHS is positive. Instead, when $x < 0$, we should have $$\frac{1}{x} = -\sqrt{\frac{1}{x^2}}.$$ So to preserve the sign in the denominator, we must write instead $$\frac{7}{\frac{1}{x}\sqrt{x^2 + 7x} - 1} = \frac{7}{-\sqrt{1 + 7/x^2} - 1},$$ and now the correct limiting behavior is retained.

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$$\lim_{x \to -\infty} \sqrt{x^2+7x}+x=$$ $$=\lim_{x \to -\infty} |x|\sqrt{1+\frac{7}{x}}+x=$$ $$=\lim_{x \to -\infty} -x\sqrt{1+\frac{7}{x}}+x=$$ $$=\lim_{x \to -\infty} -x\Big(\sqrt{1+\frac{7}{x}}-1\Big)=$$ $$=\lim_{x \to -\infty} -x\Big(1+\frac{7}{2x}+O\Big(\frac{1}{x^2}\Big)-1\Big)=$$ $$=\lim_{x \to -\infty} -\frac{7}{2}+O\Big(\frac{1}{x}\Big)=-\frac{7}{2}$$

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The other answers have already addressed how to calculate the limit, so I won't go into that detail here. Instead, this is another approach to avoid problems with negative numbers.

Since $x$ is approaching $-\infty$, let's make the substitution $u = -x$. Then the problem becomes: $$\lim_{u \to \infty} \sqrt{u^2-7u}-u$$ and it should be straightforward from here.

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  • $\begingroup$ But doesn't that become $0$ (in the denominator), so there wouldn't be a limit, which is contrary to what heropup states? (I'm new to this, so I'm sorry if I made another mistake.) $\endgroup$ – Max Li Jul 27 '16 at 18:38
  • $\begingroup$ @MaxLi The denominator will be $\sqrt{u^2-7u}+u$, which is equal to $\frac1u \left(\sqrt{1-\frac7u}+1\right)$. After cancelling the factor of $\frac1u$ with the numerator, this term will approach $2$. $\endgroup$ – Théophile Jul 27 '16 at 19:49

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