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)?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ Hmm... seems like the simplest way is best! +1 $\endgroup$ – Mathmo123 Jul 31 '14 at 22:55

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.