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I was given this problem by a friend:

$$ \def\limit{\lim_{x\to5}} \def\top{\sqrt{x}-2} g(x) = \frac{\top}{x-5}\\ \limit g(x) = \quad? $$

This caught me by surprise, because I can't remember how to do this problem with basic Calculus. Intuitively, the limit doesn't exist since $\limit(\top) \ne 0$ but $\limit(x-5) = 0$. And the limit doesn't even approach either infinity since the bottom is an odd-power polynomial.

However, how can I prove this by using basic Calculus that a new Calculus student would understand?


I tried to do this as follows, but it seems overly complicated:

I spent some time and found this function:

$$ f(x)=\frac1{\sqrt{x-5}}\\ f(x) < \frac\top{x-5}\\ \text{when }5 < x < \frac{81}{16} $$

And we know that $\lim_{x\to5^+} f(x) = \infty$, so therefore,

$$\lim_{x\to5^+} g(x) = \infty$$

However, $f(x)$ doesn't work for the LH limit. However:

$$ g(x) > 0 \quad\text{if}\quad x > 5\\ g(x) < 0 \quad\text{if}\quad 4 < x < 5 $$

So that means that $\lim_{x\to5^-} g(x) < 0 \ne \infty$ so the limit does not exist, and we can't even say that the limit is one of the infinities.

Isn't there an easier way to do this (assuming graphing isn't allowed)?

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  • $\begingroup$ Perhaps my problem is that I'm missing out on some obvious detail such as how to evaluate the RH/LH limits without comparing to another function. $\endgroup$ – Justin Jul 31 '14 at 22:41
  • $\begingroup$ btw a proof does not have to have a complicated calculation in it. In my opinion your original reasoning is good enough as a proof. $\endgroup$ – Winther Jul 31 '14 at 23:04
  • $\begingroup$ @Winther Unfortunately, the reasoning might not be enough for a Calculus teacher. It's enough to show that it is true, but it is not well-formed enough to submit to a teacher IMO, because there are usually points for work. Furthermore, beginning Calculus student's can't be expected to know that. $\endgroup$ – Justin Jul 31 '14 at 23:06
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    $\begingroup$ I understand (vagely remember those days):) Well, if you need more details/reasoning then users84413's answer below is the simplest way to go in my opinion. $\endgroup$ – Winther Jul 31 '14 at 23:10
  • $\begingroup$ Can anyone explain why we can't use L'Hopital's rule here and get $\frac{1}{2\sqrt5}$ as the limit? $\endgroup$ – Mr Reality Oct 15 '17 at 4:19
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Probably the best way to do this is your original argument:

If $\displaystyle\lim_{x\to c} f(x)\ne 0$ and $\displaystyle\lim_{x\to c} g(x)=0$, then it follows that $\displaystyle\lim_{x\to c} \frac{f(x)}{g(x)}$ does not exist, since

$\displaystyle\lim_{x\to c} \frac{f(x)}{g(x)}=L\implies\lim_{x\to c}f(x)=\lim_{x\to c} \left(\frac{f(x)}{g(x)}\right)\big(g(x)\big)=L\cdot0=0$, which gives a contradiction.

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    $\begingroup$ Hmm... seems like the simplest way is best! +1 $\endgroup$ – Mathmo123 Jul 31 '14 at 22:55
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Hint: $$\frac{\sqrt x - 2}{x-5}=\frac{(\sqrt x -\sqrt5)+(\sqrt5- 2)}{x-5}=\frac{\sqrt x -\sqrt5}{x-5}+\frac{\sqrt 5 - 2}{x-5}=\frac{\sqrt x -\sqrt5}{(\sqrt x +\sqrt 5)(\sqrt x -\sqrt 5)}+\frac{\sqrt 5 - 2}{x-5}=\frac 1 {\sqrt x+\sqrt 5}+ \frac{\sqrt 5 - 2}{x-5}$$


This kind of trick is really useful and happens often. It is easiest to illustrate how to use it on a simple example, $\frac{x}{x+1}$. The aim is to find something where adding and subtracting it will enable us to cancel the denominator. In this case, we notice that adding and subtracting $1$ will do the trick: $$\frac{x}{x+1}=\frac{(x+1) -1}{x+1}=1-\frac{1}{x+1}$$In the above case, spotting that $x-5 = (\sqrt x +\sqrt 5)(\sqrt x -\sqrt 5)$ leads to the choice of adding and subtracting $\sqrt 5$ and lose one of the factors in the denominator.

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  • $\begingroup$ Nice, this makes sense. My only question is how did you figure out to add/subtract $\sqrt5$? $\endgroup$ – Justin Jul 31 '14 at 22:45
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    $\begingroup$ Simply because $$\lim_{x\to 5}\frac{\sqrt{x}-\sqrt{5}}{x-5}$$ is the definition of the derivative of $\sqrt{x}$ at $x=5$. @Quincunx $\endgroup$ – Thomas Andrews Jul 31 '14 at 22:46
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    $\begingroup$ I've added more detail - this is a trick that comes up all over the place and will save you a lot of time in the future... it takes a bit of practice to get used to though. $\endgroup$ – Mathmo123 Jul 31 '14 at 22:50
  • $\begingroup$ @Quincunx the easier way to do this is to factorise the denominator as a difference of two squares. I've edited this in the answer. $\endgroup$ – Mathmo123 Jul 31 '14 at 22:53
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Distinguish between the two cases:

$\lim_{x \to 5^{\color{blue}{-}}} \frac{\sqrt x -2}{x-5}$ and

$\lim_{x \to 5^{\color{blue}{+}}} \frac{\sqrt x -2}{x-5}$

These limits can be easily evaluated. When $x \to 5^{\color{blue}{-}}$, the denominator goes to 0 and is negative while the numerator doesn't/isn't, leading the fraction to go to $-\infty$. Similarly for the other side.

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  • $\begingroup$ And how does one evaluate each side? I evaluated the second using a complicated method, perhaps there is an easier way? $\endgroup$ – Justin Jul 31 '14 at 22:48
  • $\begingroup$ If $x \to 5^-$ then the denominator is negative and goes to 0. Thus the fraction goes to $-\infty$ With the same argumentation you can find the limit for $x \to 5^+$ $\endgroup$ – callculus Jul 31 '14 at 22:54
  • $\begingroup$ Please edit that into your answer. Simply that when $x\to5^-$, the denominator goes to 0 and is negative and the numerator doesn't/isn't, leading the fraction to go to $\-infty$, and similarly for the other side. $\endgroup$ – Justin Jul 31 '14 at 22:58
  • $\begingroup$ It is possible, that the guys, who downvoted my answer post a reason ? Thanks. $\endgroup$ – callculus Jul 31 '14 at 23:06
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    $\begingroup$ I did not downvote your post, but I believe the downvote is because in the work I showed in the OP, I clearly distinguish between the cases. You are thinking even more than that, because you know how to evaluate those two limits. I recommend that you include in your answer the way to evaluate those limits. $\endgroup$ – Justin Jul 31 '14 at 23:08

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