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 $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$

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



However, I multiplied

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

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



Which is undefined.

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

  • 3
    $\begingroup$ $-1\neq \sqrt{(-1)^2}$. 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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ Wow, that's a lot of tricky business to watch out for. Thanks. $\endgroup$ – Max Li Jul 27 '16 at 18:40
  • 1
    $\begingroup$ 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

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.


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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.