8
$\begingroup$

I know that when l'Hospital's rule is taken for the $\frac{0}{0}$ case, we get that the functions start looking at lines, like in this figure from Stewart's Calculus:

But what is the visual when the limit is the $\frac{\infty}{\infty}$ case?

The only thing that comes to mind that is easily visualized is a rational function's asymptotes (and that comes from dividing numerator and denominator by the largest exponent in the denominator)

$\endgroup$
5
  • 5
    $\begingroup$ $\frac{f}{g} = \frac{1/g}{1/f}$ so you can apply the $\frac{0}{0}$ case to the functions $1/f$ and $1/g$. $\endgroup$ – Jaap Scherphuis Jan 19 at 8:22
  • $\begingroup$ Your visual in question is not actually about LHospital but about finding limit of $f/g$ at $c$ when $f(c) =g(c) =0$ and $f, g$ are differentiable at $c$ with $g'(c) \neq 0$. The limit in this case is evaluated using definition of derivative. $\endgroup$ – Paramanand Singh Jan 19 at 10:56
  • $\begingroup$ A visual for LHospital actually consists of a curve passing through origin and analyzing parametric equation of the curve. $\endgroup$ – Paramanand Singh Jan 19 at 10:57
  • $\begingroup$ @JaapScherphuis: Hmm. Really? It is most definitely not this simple. Note that $\frac FG = \frac{1/G}{1/F}$, but applying L'Hôpital here gives a mess: $$\frac{-G'/G^2}{-F'/F^2}$$ doesn't seem to resemble $\frac{F'}{G'}$. $\endgroup$ – Ted Shifrin Jan 27 at 20:23
  • $\begingroup$ @ParamanandSingh Would the geometric interpretation on this site be better? math.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/… $\endgroup$ – user612996 May 19 at 2:46
2
+100
$\begingroup$

Well, you actually CAN do it.

There is one simple way, which sounds a bit like cheating: taking the limit means looking at the function in a neighbourhood of infinity, so you could just end your graph at a really big number and see how the slope behaves.

However, there is also a more insightful point of view, but it is really long and pretty much useless. Once you have the right intuition for the finite case, you just use that.

The idea for what you're asking would be "squeezing" the real line $\mathbb{R}$ to an interval. For example: $atg([0,\infty))=[0,\frac{\pi}{2})$, where you could consider the closed interval $[0,\frac{\pi}{2}]$ regarding $\frac{\pi}{2}$ as the image of $\infty$.

Then you could draw your function using these coordinates: what you get is a deformed & squeezed graph which is contained in the square $[0,\frac{\pi}{2}]\times[0,\frac{\pi}{2}]$.

Meanwhile, you should also write down the rule for how to convert a slope according to the squeezed graph into the right slope (i.e. according to the original graph).

Finally, you should be able to see that the slope given by the squeezed graph is actually that given by De L'hopital's rule (adequately transformed).

This is a little example I tried to carry out on the spot: here is the squeezed graph of the function $x^2$ (I'm thinking about computing the limit $\lim_{x\rightarrow \infty} \frac{x^2}{x}$, which should obviously be $\infty$, but I wish to use De L'hopital for understanding purpose)

You can see that at infinity (the upper right part) the slope is zero. the transformation rule for slope of this change of variables is probably something like

$ real\_slope = \frac{1}{squeezed\_slope} $

(I din't check it though).

So you correctly get that the real slope is $\infty$ as you would expect by applying De L'hopital's rule.

As I said this is long and not necessary. Also, beware that I omitted the details and you should go check them, if you want a full understanding. I actually never got further than this, cause I didn't feel the urge. I also had the same question and this is an answer I gave myself. If you like this kind of questions, you should like measure theory.

$\endgroup$
0
$\begingroup$

L'Hospital is based on the fact that a continous function can be locally approximated by its tangent. If you want to evaluate the limit at infinity, for example $\lim \limits_{x \to \infty} \frac{\sqrt{x}}{\ln{x}}$, the tangents would also lie there so there is no visual for that.

$\endgroup$
1
  • $\begingroup$ Except you perform the transformation $$x\mapsto \frac 1x.$$ $\endgroup$ – Allawonder Jan 27 at 21:38
0
$\begingroup$

I would rather split it into four cases:

  1. limit of type $\frac00$ in finite point
  2. limit of type $\frac00$ in infinity
  3. limit of type $\frac\infty\infty$ in finite point
  4. limit of type $\frac\infty\infty$ in infinity

The case 2 doesn't seem to be covered by those ilustrations either. I thought how it could work for case 4:

Use the same illustrations but move the point $a$ to $0$. Then you can see that the limit is equal to the ratio of the slopes of the functions approximated by lines.

$\endgroup$
1
  • $\begingroup$ In case 4., if $\lim_{x\to\infty} f'(x)=L$ and $\lim_{x\to\infty} g'(x)=M$, then you can make the argument that for large $x$, we have (for suitably large $a$) $f(x)\approx L(x-a)+f(a)$ and $g(x)\approx M(x-a)+g(a)$, and then it works. $\endgroup$ – Ted Shifrin Jan 27 at 20:26
0
$\begingroup$

The counterpart of 0/0 to infinities can be seen graphed in the example below. Although numerator and denominator individually must necessarily go to $\infty,$ the quotient goes to a constant in the limit, in this case to 1.

enter image description here

Applying L'Hospital's Rule for any $(a,b)$ values

$$ lim_{x\to \infty}\dfrac {\log(a+3x)}{\log(b+x)} = \dfrac31\dfrac{b+x}{a+3x}=\dfrac31\dfrac{b/x+1}{a/x+3}\to 1 $$

$\endgroup$
3
  • $\begingroup$ It is most definitely not this simple. Note that $\frac FG = \frac{1/G}{1/F}$, but applying L'Hôpital here gives a mess: $$\frac{-G'/G^2}{-F'/F^2}$$ doesn't seem to resemble $\frac{F'}{G'}$. $\endgroup$ – Ted Shifrin Jan 27 at 20:23
  • $\begingroup$ Answer changed to show better what I meant. $\endgroup$ – Narasimham Jan 28 at 22:55
  • $\begingroup$ I don't see that this is relevant to the question anymore. $\endgroup$ – Ted Shifrin Jan 28 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy