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I was doing some work on limits for calculus: rudimentary exercises in analyzing a limit for a given expression.

The problem that came up was this one: $$\lim_{x\to 1}\frac{\sqrt{x+5}+\sqrt6}x\tag1$$

I could immediately plug in the limit, and find the result as: $$\lim_{x\to 1} 2\sqrt6\tag2$$

However, at first I conjugated, because I assumed the limit was a zero like the others, and this showed up: $$\lim_{x\to1}\frac{x-1}{x(\sqrt{x+5}-\sqrt6)}\tag3$$

Now, the limit for that one does not exist. Why does this happen? It would seem that conjugating would not affect the final result being different since I am just manipulating the expression, yet it did, and yet it is not incorrect to apply a conjugate here, it seems. Why would it be incorrect to apply a conjugate here, then?

Blue
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BigRigz
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    All those limits exist, and are equal. Just the function in the last one is not defined at $x=1.$ – Thomas Andrews Jul 04 '21 at 23:01
  • That sounds confusing to me—at least, me not as a mathematician. There is only one limit indicated here, and how can a function be defined in terms of a single domain point? That is not how a function works, I think. – BigRigz Jul 04 '21 at 23:04
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    A simpler example is the difference between $$\lim_{x\to 1}\frac1x\\text{and}\ \lim_{x\to1}\frac{x-1}{x^2-x}$$ – Thomas Andrews Jul 04 '21 at 23:07
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    Who said a function was “defined in terms of a single point.” The limit of a function $f(x)$ as $x\to 1$ is about the values of $f(x)$ when $x$ is around $1,$ but does not depend on the value of $f(1)$ or even if $f(1)$ is defined. – Thomas Andrews Jul 04 '21 at 23:09
  • @ThomasAndrews Then I misunderstood your initial statement. I am thinking in terms of someone who is just starting to learn calculus, and it seems weird that the function that does not have a limit defined at x = 1 would simply mean the limit is elsewhere. Maybe that is related to the epsilon-delta rule, but I myself still do not know how that functions either. Perhaps I am not ready to understand the answer to this question. – BigRigz Jul 04 '21 at 23:13
  • @BigRigz "at first I conjugated" — How exactly did you do that? You probably multiplied both numerator and denominator by $;\sqrt{x+5}-\sqrt6;$ *but* you are only allowed to do that when the multiplicand is non-zero i.e. $\sqrt{x+5}-\sqrt6 \ne 0 \iff x \ne 1$. Then of course you can no longer calculate the limit for $x \to 1$ by directly substituting $x=1$ in the resulting expression, since that expression is not valid when $x=1$. – dxiv Jul 04 '21 at 23:30
  • I'm worried by your saying "I could immediately plug in the limit", which suggests that maybe you're conflating "limit" and "plug in". Some limits (like the original one in your question) can be evaluated by plugging in, but many cannot. In particular, after you multiplied numerator and denominator by that conjugate, you got an expression where plugging in produced nonsense ($0/0$). Nevertheless, the limit exists (as @ThomasAndrews said). You should get used to limits that are perfectly OK even though plugging in gives $0/0$, because such limits play a key role in calculus (derivatives). – Andreas Blass Jul 04 '21 at 23:39
  • I am, alas, a layman, and can not seem to get the essence of what you all are trying to say: I am just starting to learn this! And, I see that there is a lot that I need to learn to begin to understand why what I did is wrong, but I say again, I am just starting! However, you all are providing strong guidance that I just need to continue on, and eventually these concepts you are highlighting will be clear to me. I would like to thank you all for the guidance, but perhaps we can put this to a close for me to reflect on, lest I make you more mad with my nonsense. – BigRigz Jul 04 '21 at 23:44
  • The main thing that seems to be troubling you is that "the limit of the expression as $x\to 1$" is a different concept from "the value of the expression when $x=1$". They have different notations and different words because they are not the same thing! Your confusion is not uncommon, but you should try as hard as you can to put it behind you and understand how they differ, or else confusion about the difference will keep making you reach wrong conclusions. – Troposphere Jul 05 '21 at 00:52
  • That would make sense, given that I am still thinking in terms of algebra, and not the newer rules of calculus still unclear to me. – BigRigz Jul 05 '21 at 00:53

2 Answers2

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The function $f_1(x) = \frac{\sqrt{x + 5} + \sqrt{6}}{x}$ is defined for all $x \geq -5$ except for $x = 0$, and in fact is continuous on that domain, and so $\lim_{x \rightarrow 1} f_1(x) = f_1(1) = 2\sqrt{6}$.

The function $f_2(x) = \frac{x - 1}{x(\sqrt{x+5} - \sqrt{6})}$ is defined for all $x \geq -5$ except for $x = 0$ and $x = 1$, because when $x = 1$ it evaluates to $\frac{0}{0}$.

However, it is continuous everywhere on its domain, and in particular $\lim_{x \rightarrow 1^-} f_2(x) = \lim_{x \rightarrow 1^+} f_2(x)$, i.e. the left- and right-hand limits approaching $x = 1$ are the same, so in fact we can say that $\lim_{x \rightarrow 1} f_2(x) = 2 \sqrt{6}$. This is a removable discontinuity - a point where the function is not defined but where you can define a value that makes the function continuous again.

It's the same as the behaviour of $\frac{x}{x}$ as $x \rightarrow 0$ - the function is undefined, but it has a well-behaved limit that lets you fill in the missing spot.

Your mistake came because when you tried using the conjugate method, you multiplied your function by $\frac{\sqrt{x+5} - \sqrt{6}}{\sqrt{x+5} - \sqrt{6}}$, which at $x = 1$ is the same as multiplying by $\frac{0}{0}$ - you were the one who introduced the discontinuity there. The whole point of taking the limit is to avoid the discontinuity, which you can do as long as the function is well-behaved close to where it's discontinuous.

ConMan
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  • I would be careful with your usage of the word "range". Just FYI. – Cameron Williams Jul 04 '21 at 23:34
  • Thanks, had a small brain fart when I meant to say domain, will fix. – ConMan Jul 04 '21 at 23:35
  • When I approach further concepts, perhaps this will make more sense, though your use of fundamental words here (or my notion of what often appears in a discussion on calculus) indicate that it will make sense pretty soon. Thank you. – BigRigz Jul 04 '21 at 23:39
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It is enough to evaluate the function in 1, since it does not generate any indeterminacy or expression that does not exist.

  • This doesn't seem to answer the OP's confusion about what goes wrong (he thinks) in the rewriting that introduces a removable singularity. – Troposphere Jul 05 '21 at 00:48