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This question already has an answer here:

don't get mad at how bad this looks, I'm trying my best. Anyways, I was looking at the definition of $e$, you know like: $$\lim_{x\to\infty}\left(1+\frac1 x\right)^x$$

but if I use the properties of limits I can take the exponent outside the limit $$\left(\lim_{x\to\infty}1+\frac1 x\right)^x$$

then if I use direct substitution, the limit of $1/x$ as $x$ approaches infinity is $0$ and the limit of $1$ as $x$ approaches infinity is just $1$ so now the limit is:

$$1^\infty$$

and that just equals $1$, so now it seems like $e$ just equals $1$. What's wrong with my math or with what I just did?

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marked as duplicate by Guy Fsone, J. M. is a poor mathematician, Claude Leibovici calculus Nov 12 '17 at 6:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See here for a basic tutorial on MathJax. $\endgroup$ – d80d2729a352b1366139fc119d3345 Jun 12 '14 at 15:32
  • $\begingroup$ are you sure you can just take the exponent outside the limit, if the exponent is the variable of the limit? $\endgroup$ – cirpis Jun 12 '14 at 15:33
  • $\begingroup$ You cannot "use direct substitution". "$\lim_{x\to\infty} g_x(x)\neq \lim_{x\to\infty} g_\infty(x)$" $\endgroup$ – Clement C. Jun 12 '14 at 15:33
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    $\begingroup$ I think you are referring to this property. Let $\lim{f(x)}$ exits and $a$ be a $\color{red}{\text{real constant}}$, if $\lim{(f(x)}^a)$ exists then $\lim{(f(x)^a)}=(\lim{f(x)})^a$. $\endgroup$ – d80d2729a352b1366139fc119d3345 Jun 12 '14 at 15:36
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    $\begingroup$ Others have already told you you can't take a bound variable outside the limit, but even then, the form $1^\infty$ is still ill-defined, so the argument doesn't hold water either way. $\endgroup$ – mval Jun 12 '14 at 15:40
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You can tell this is wrong by using the binomial theorem for a positive integer index and making $x$ a positive integer $N$, so that $$\left (1+\frac 1N\right)^N=1^N+\binom N11^{N-1}\left(\frac 1N\right)^1+\dots=2+\dots$$Where there is a finite number of omitted terms, all positive. Whence the limit, if it exists, must be $\ge 2$. Further analysis of this expansion gives the correct limit with a little work.

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There are two main errors:

the first one is that you can't just take the limit inside because the exponential is itself a function of $x$; even intuitively I believe it's clear that is wrong.

Plus, $1^{\infty}$ is not $1$.. You can say that $\lim_{x \to \infty} 1^x = 1$, but your $(1 + \frac1x)$ is not exactly $1$, it just goes to $1$. This means that you can't say that it's limit is $1$ if it is being raised by something that $\to \infty$

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  • $\begingroup$ I also believe that this is wrong, that's why I am asking this because I just looked at this one day and decided to try and expand it to see what happens and this is what I found. The main objection that people are giving is that you can't just take the exponent out of the function, is that just because the exponent is x? and with the lim x-->infinity 1 + 1/x you can use the property where you take the limit of each individually and that's where I got 1, but that hasn't really been a big objection to the false proof. $\endgroup$ – eric Jun 12 '14 at 19:07
  • $\begingroup$ @eric The first wrong step is taking the limit inside, because the exponent is $x$. I pointed out that also the second step is wrong; granted this has less importance because the proof is already invalidated $\endgroup$ – Ant Jun 12 '14 at 19:10

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