2
$\begingroup$

First some context:

I was trying to find the limit of $\frac{e^x - 1}{x}$ as x approaches zero without using L'Hopital's rule to avoid circular reasoning.

Then, I was told that I could use the definition:

$$e = \lim_{n\to\infty} \left( 1+\frac{1}{n} \right)^n$$ The trick is to set $n = \frac{1}{x}$

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

So, my limit becomes: $$\lim_{x\to 0} \frac{1+x-1}{x} = 1$$

This makes sense on an intuitive level. By setting n equal to $\frac{1}{x}$, I am approaching the limit from a specific direction on a two-dimensional plane. However, I have not worked with two-dimensional limits before. When is this allowed? Surely there must be situations in which this does not work because the limit is different depending upon what path you use to approach the point at which you are taking the limit. How do you test if this "trick" gives the correct answer?

$\endgroup$
  • $\begingroup$ you're not actually approaching in a 'two-dimensional plane' (you're conflating the change of variables with some sort of limit along a path given by the graph of $n\to 1/n$) but instead still allowing $n$ to grow unbounded in a continuous way, only in a qualitatively different way then you might be used to to imagining it $\endgroup$ – oldrinb Jun 29 '15 at 15:15
  • 2
    $\begingroup$ Why do you think you're actually solving a two-dimensional limit by following a specific direction? Are you thinking of $x$ and $n$ as the coordinates on a plane? $\endgroup$ – Giorgio Comitini Jun 29 '15 at 15:16
  • $\begingroup$ @Giorgio Comitini Yes, that is how I think about this trick in my attempt to make sense of it. $\endgroup$ – Avatrin Jun 29 '15 at 16:04
  • $\begingroup$ Well, I'm not sure that is the correct way of thinking about your solution (which, by the way, is totally legal). What you're really doing is first of all passing from the limit of a sequence to the limit of a function (i.e. $n\mapsto y$, with $y\in]-\infty,+\infty[$ such that $y|_{\mathbb{N}}=n$) and then computing the limit with respect to $x(y)=\frac{1}{y}$. Both these steps are completely well-defined in one-variable calculus (they are already well defined in topology). $\endgroup$ – Giorgio Comitini Jun 29 '15 at 16:10
  • $\begingroup$ Well, I am asking how I can know whether or not this method is valid. Is there a book you recommend to learn this? I have had real analysis, but I never encountered this. $\endgroup$ – Avatrin Jun 29 '15 at 22:48
2
$\begingroup$

To rephrase, OP asks about evaluating $$\lim_{x\to 0}\lim_{n\to \infty} \frac{ \left(1+\frac{1}{n}\right)^{nx}-1}{x}$$ along the path $n=\frac{1}{x}$. Equivalently, setting $n=\frac{1}{y}$, the limit becomes $$\lim_{x\to 0}\lim_{y\to 0} \frac{ \left(1+y\right)^{x/y}-1}{x}$$

OP is quite right to be concerned about this method. If the limit $$\lim_{(x,y)\to (0,0)} \frac{ \left(1+y\right)^{x/y}-1}{x}$$ exists, then it is the same along every path. The specified path $(x=y)$ provides a slick way of determining what that limit must be, if it exists at all. However there are many similar expressions where the limit does not exist; to use the path given one must prove that the limit exists at all.

$\endgroup$
  • $\begingroup$ Yes, this is exactly what I am asking about. Thank you for phrasing it better than I was able to! $\endgroup$ – Avatrin Jun 29 '15 at 16:07
  • $\begingroup$ What methods can I use to prove that the limit exists? Is there maybe some book you can recommend for me to read to learn this? $\endgroup$ – Avatrin Jun 29 '15 at 22:52
  • 1
    $\begingroup$ This is a calc 3 - type question. $\endgroup$ – vadim123 Jun 29 '15 at 23:19

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.