$e = 1$? something is wrong here: $\lim_{x\to\infty}\left(1+\frac1 x\right)^x$=? [duplicate]

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?

marked as duplicate by Guy Fsone, J. M. is a poor mathematician, Claude Leibovici calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 12 '17 at 6:40

• See here for a basic tutorial on MathJax. – d80d2729a352b1366139fc119d3345 Jun 12 '14 at 15:32
• are you sure you can just take the exponent outside the limit, if the exponent is the variable of the limit? – cirpis Jun 12 '14 at 15:33
• You cannot "use direct substitution". "$\lim_{x\to\infty} g_x(x)\neq \lim_{x\to\infty} g_\infty(x)$" – Clement C. Jun 12 '14 at 15:33
• 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 $\lim{(f(x)^a)}=(\lim{f(x)})^a$. – d80d2729a352b1366139fc119d3345 Jun 12 '14 at 15:36
• 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. – mval Jun 12 '14 at 15:40

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.
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 $\to \infty$
• @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 – Ant Jun 12 '14 at 19:10