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Is this limit defined or undefined? $$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$ When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it's defined on the right side.

By multiplying by $\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$: $$\lim\limits_{x \to 0+} \frac{\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} \, \right) \left(\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}} \, \right)}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$

$$=\lim\limits_{x \to 0+} \frac{\frac{1}{x}+2-\frac{1}{x}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$ $$=\lim\limits_{x \to 0+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$ Then, we multiply by $\frac{\sqrt{x}}{\sqrt{x}}$: $$=\lim\limits_{x \to 0} \frac{2\sqrt{x}}{\sqrt{1+2x}+1}$$ And, we substitute: $$=\frac{2\sqrt{0}}{\sqrt{1+2\times0}+1} = 0$$ So, is this limit defined or not? and what's my error, if any?

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    $\begingroup$ Notice that you can only break a limit up if each limit you're creating is finite. So you can't split the limit up here. In general, any limit law needs to be done with only finite limits involved. $\endgroup$ – Kaynex Nov 11 '17 at 2:37
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    $\begingroup$ Your solution is fine. Certainly the limit of a difference can exist even if the limits of the terms being subtracted do not exist individually. Consider $\lim_{x \to \infty} (x - x)$, for example. $\endgroup$ – Bungo Nov 11 '17 at 2:40
  • $\begingroup$ $\infty - \infty$ is not in the domain of subtraction, so the fact subtraction is continuous does not apply here. $\endgroup$ – Excluded and Offended Nov 11 '17 at 2:52
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    $\begingroup$ Your title is wrong: this is not the difference of two undefined limits (count the number of times $\lim$ occurs in your formula). If it were, it would definitely be undefined: arithmetic assumes definite operands. $\endgroup$ – Marc van Leeuwen Nov 11 '17 at 10:27
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    $\begingroup$ A simple example of two expressions whose limits are undefined but the limit of the difference is clearly defined: take $a_n = b_n = (-1)^n.$ Then both $\lim_{n\to\infty} a_n$ and $\lim_{n\to\infty} b_n$ are undefined but $\lim_{n\to\infty} (a_n-b_n) = 0.$ $\endgroup$ – md2perpe Nov 11 '17 at 10:45
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What you did is correct. The point is that in its initial form, your problem was of an indeterminate form. Basically, if we have a limit that "looks like" $\infty-\infty$ or something like this, the value can basically be anything under the sun. It can be illuminating to see the process in reverse. $$ 0=\lim_{x\to\infty} 0=\lim_{x\to\infty}(x-x)\ne \lim_{x\to\infty} x-\lim_{x\to\infty}x"="\:\text{nonsense}.$$

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Remember that the rule that you referred to, "the rule of difference of limits", is not just the equation $$ \lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x) $$ but rather the statement that, if both of the limits on the right side of this equation are real numbers, then the limit on the left side (is also a real number and) is given by this equation. So this rule does not apply to the limit in your question.

More generally, when learning rules (or theorems or principles or whatever they may be called), don't just learn formulas, but pay attention also to the words around them. The words are not just decoration but are essential for the correctness of the rule.

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    $\begingroup$ If students really pay attention to your last paragraph, most of such questions about mathematics would simply vanish. =) But sadly, I've seen many textbooks that don't even have any of the necessary words to make their 'rules' correct... $\endgroup$ – user21820 Nov 11 '17 at 9:30
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    $\begingroup$ @user21820 So maybe that paragraph in my answer should be addressed not only to students learning the rules but also to textbook authors writing the rules (who surely should know better). $\endgroup$ – Andreas Blass Nov 11 '17 at 15:37
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    $\begingroup$ (+1) And I'd like to give you and additional +1 just for "The words are not just decoration but are essential for the correctness of the rule.". I think I'd steal your expression and "recycle" it for my students that blindly apply formulas without even thinking about the meaning of the variables they contain! :-) $\endgroup$ – Lorenzo Donati Nov 11 '17 at 21:14
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Just to add to what has been said, a limit expression that has undefined (or infinite) value cannot be treated as an ordinary real expression, and so it technically is invalid (senseless) to manipulate it as if it were a real number. For example $\sin(n)-\sin(n) \to 0$ as $n \to \infty$, but "$\lim_{n\to\infty} \sin(n)$" itself is simply undefined and it is technically invalid to even write "$\lim_{n\to\infty} \sin(n) - \lim_{n\to\infty} \sin(n)$", not to say ask for its value (unless you want to have propagation of undefined values...).

So the literal answer to your question is:

The difference of 2 undefined limits cannot be defined, by definition. (Even if you wish to permit writing potentially undefined expressions, it would not make a difference, since any expression with an undefined subexpression will itself be undefined.)

The correct statement is that it is possible for the difference of two expressions to have a limit even though neither of the expressions has a limit (under the same limiting conditions).

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I have the same take as user21820. Your error is treating “undefined” like it's a value when it's really just a predicate.

When we write $\lim_{x\to a} f(x) = L$, it's not really an equation so much as a statement. It's a statement about $f$, $a$, and $L$ all in one. When we say that $\lim_{x\to a} f(x)$ is undefined, we mean that $\lim_{x\to a} f(x) = L$ is not true for any number $L$.

It's tempting to think of “undefined” as some magic quantity which nullifies real numbers. Sometimes this leads to true statements. For instance, if $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x)$ is undefined, then $\lim_{x\to a} (f(x) + g(x))$ is undefined. You might want to think of this succinctly as “finite plus undefined equals undefined.”

But it also leads to false statements, as you have discovered. It's just not true that if $\lim_{x\to a}f(x)$ is undefined and $\lim_{x\to a}g(x)$ is undefined, then $\lim_{x\to a} (f(x) + g(x))$ is undefined. The simplest counterexample would be if $g(x) = -f(x)$. So even though you might want to think to yourself “undefined plus undefined equals undefined,” this is specious reasoning.

In a lot of computer languages, you can have an undefined or null value, and in some cases it can combine with other values. For instance, in Excel, when a formula in a cell evaluates to #NA, any formula that uses that cell will also evaluate to #NA. But it doesn't work that way with limits in math.

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You write $x\to 0+$ but it's often written as $x\to 0^+$. You asked if your limit is defined or undefined. I'll answer that with another two ways.

$$\lim_{x\to 0^+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)=$$

$$=\lim_{x\to 0^+}\left(\frac{\sqrt{1+2x}-1}{x}\cdot \sqrt{x}\right)=$$

Use the definition of a derivative and a limit multiplication rule/law.

$$=(\sqrt{1+2x})\bigg|_{x=0^+}\cdot 0=0,$$

because the derivative is a real number.

Another way is using Newton's generalized binomial theorem, also see Binomial series, which converges when $|x|<1$ but it can diverge when $|x|\ge 1$, see the Binomial series link for more information. $$\lim_{x\to 0^+}\frac{\sqrt{1+2x}-1}{x}=$$

$$=\lim_{x\to 0^+}\frac{(1+2x)^{\frac{1}{2}}-1}{x}=$$

$$=\lim_{x\to 0^+}\frac{1+\frac{1}{2}2x+o(x)-1}{x}=$$

$$=\lim_{x\to 0^+}\left(1+\frac{o(x)}{x}\right)=1+0=1$$

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What you wrote,$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}},$$is undefined, because it is the difference of$$\lim\limits_{x \to 0+} \sqrt{\frac{1}{x}+2},$$which is undefined, and a another quantity$$\sqrt{\frac{1}{x}},$$the latter being meaningful, although dependent on the variable $x$. What I think you meant to write is $$\lim\limits_{x \to 0+}\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right).$$ This is not, as the title of your question says, "the difference of two undefined limits". Rather, it is a (single) limit of a quantity (that happens to be expressed as a difference). As long as you treat this as a single limit, which is what you have done quite correctly, no problem arises.

In general, it is a false move to split a limit of a difference into the difference of two limits. It works sometimes, but then it needs to be justified carefully. On the other hand, if two limits are separately well defined, then their difference can be expressed as the limit of a difference (assuming that the variable, domain, and endpoint of the limiting process is the same for both).

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    $\begingroup$ Come on. What you are saying is that large operators like $\lim$ (or $\sum$) have operator precedence higher than addition and subtraction and that therefore the formula needs parentheses. This is indeed the common convention (also the precedence is lower than multiplication; not sure about division, as even $ab/cd$ might give discussion), but if that is what you want to say, then just say directly. $\endgroup$ – Marc van Leeuwen Nov 11 '17 at 10:33
  • $\begingroup$ @MarcvanLeeuwen: You are of course quite correct about precedence. However, given the OP's muddle about whether one or two limits were involved, I thought it appropriate to treat this concrete case in detail rather than state a general rule (which might be misunderstood or dismissed as stylistic pettiness). $\endgroup$ – John Bentin Nov 11 '17 at 10:58
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    $\begingroup$ I have my nitpicks about notation, too, that I am determined to bring up whenever I see them. But the right place to do so is in a comment. In a particularly egregious case, consider editing the question/answer and append a comment explaining why. $\endgroup$ – Michael Grant Nov 11 '17 at 14:51

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