# Which answer is correct? Finding the limit of a radical as $x$ approaches infinity.

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?

• $-1\neq \sqrt{(-1)^2}$. The square root returns the principle square root, which in the case of non-negative reals is always non-negative. Commented Jul 27, 2016 at 18:00
• @JMoravitz it's not. I want to know specifically about its application to limits in this case, and which is correct. Commented Jul 27, 2016 at 18:03
• The application to limits is "don't make an invalid simplification". ​ ​
– user57159
Commented Jul 27, 2016 at 18:04
• @RickyDemer Would you mind writing an answer saying which answer is correct? Commented Jul 27, 2016 at 18:05
• 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, $\sqrt{x^2}=|x| = \begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$ Commented Jul 27, 2016 at 18:32

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.

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.

• Wow, that's a lot of tricky business to watch out for. Thanks. Commented Jul 27, 2016 at 18:40
• You're very welcome. Also, see related approaches here, here, here, and here. Commented Jul 27, 2016 at 18:40
• You are my hero Commented Jul 27, 2016 at 18:46
• 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. ​ ​
– user57159
Commented Jul 28, 2016 at 10:46
• @Ricky: Fair enough. Commented Jul 28, 2016 at 11:43

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