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What I mean to say is, we know that:

$$\lim_{x\to\infty} \biggl(1+\frac{1}{x}\biggr)^x = e$$

But,

$$\lim_{x\to\infty} \biggl(\frac{1}{x}\biggr) = 0$$

Thus,

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

Finally, $1^{\infty} = e$.

Why is this incorrect? Can anyone explain where the math is wrong?

  • 12
    How do you justify $\lim (1+\frac{1}{x})^x = (1+\lim \frac{1}{x} )^{\lim x}$? – 2017-01-02
  • 5
    That's like saying, why is $\frac00$ undefined, when we know that $\lim_{x\to0}\frac{3x}{5x}=\frac35?$ – 2017-01-02
  • 3
    What is $\lim_{x\to\infty}(1+\frac1x)^{2x}$? That's $1^\infty$ too, right? – 2017-01-02
  • 1
    The form $1^\infty$ becomes **from a limit**, and it is indeterminated by experience: there are a lot of limits with a form $1^\infty$ that, when clearly solved, have different values, so we cant conclude the value of $1^\infty$ uniquely. – 2017-01-02
  • 3
    Your math is wrong at assuming $\lim f(g(x)) = \lim f(\lim g(x))$. – 2017-01-02

6 Answers 6

15

It's a common mistake to do a limit inside a limit like that. Yes, $\lim_{x \to \infty}\frac{1}{x} = 0$, but that tells you nothing about $\lim_{x \to \infty}(1 + \frac{1}{x})^x$ - for example, $\lim_{x\to\infty}x\cdot\frac{1}{x} = \lim_{x\to\infty}\frac{x}{x} = \lim_{x\to\infty}1 = 1$, even though it looks like you can say $\lim_{x\to\infty}x\cdot\frac{1}{x} = \lim_{x\to\infty}x\cdot 0 = \lim_{x\to\infty}0 = 0$. The key thing is that saying something like "$\lim_{x \to \infty}\frac{1}{x} = 0$" isn't saying "$\frac{1}{x}$ eventually becomes zero", it's saying "$\frac{1}{x}$ eventually gets as close as you like to zero". So, for large enough $x$, $1 + \frac{1}{x}$ is very close to $1$. But when $x$ is enormous, taking a number very close to $1$ and raising it to the $x$th power takes that "closeness" and pulls it wide open - $1.0001^{100000000}$ is so large Google refuses to calculate it. It so happens that the "smallness" of $\frac{1}{x}$ and the "bigness" of raising something to the $x$th power balance out in just the right way so that $\lim_{x \to \infty}(1 + \frac{1}{x})^x = e$. But change anything and you upset the balance - $\lim_{x \to \infty}(1 + \frac{1}{2x})^x = \sqrt{e}$, while $\lim_{x \to \infty}(1 + \frac{1}{x})^{2x} = e^2$.

The point is, you can never just replace something inside a limit with its limit, unless you're deleting the limit. That's an extremely dangerous operation which will only yield the right answer if you got obscenely lucky - because when you do that, you erase all of the information about how fast the expression approaches its limit, so you sacrifice the opportunity to balance things out.

  • 0
    Very nicely explained – 2017-01-02
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Fundamentally, the problem is that you're treating $\infty$ as though it's a real number when it's really not.

3

$1^\infty$ is indeterminate form. So $1^\infty=e$ may be true, but it may also happen that $1^\infty\ne e$

For example take $\alpha>0$ a real number and write:

$$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^{\alpha x}=e^\alpha$$

So you might as well conclude that $1^\infty=e^\alpha$

  • 0
    How is that an indeterminate form? Wouldn't 1^x as x goes to infinity just be 1? – 2017-01-02
  • 0
    No its indeterminate because it can take on many forms – 2017-01-02
  • 2
    We should not write "$1^\infty=e$" when we really mean that $1^\infty$ is an indeterminate form of a limit tending to $e$. – 2017-01-02
  • 0
    I am writing "$1^\infty=e$" in the context of [arithmetic operations on extended real number line](https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations). And the meaning is in the context of the limits is "if $f(x)\rightarrow 1$ and $g(x)\rightarrow\infty$ then $f(x)^{g(x)}\rightarrow e$", which might or might not be true, as explained in the answer. Contrast it with $2^\infty=\infty$ which is not undeterminate ("if $f(x)\rightarrow 2$ and $g(x)\rightarrow\infty$ then $f(x)^{g(x)}\rightarrow \infty$" is always true) – 2017-01-02
3

When we state that $1^\infty$ is an indeterminate form, we mean that if $f$, $g$ are two real valued functions such that $x_0$ is in the intersection of their domains and $$\lim_{x\to x_0}f(x)=1 \qquad \lim_{x\to x_0}g(x)=+\infty \tag{1}$$ then the limit $$\lim_{x\to x_0}f(x)^{g(x)}$$ is not uniquely defined by the condition $(1)$.

In the following example, we show that for every $a>0$ there exist functions $f_a, g_a$ and $x_0\in \mathbb {\tilde R}$ satisfying $(1)$ such that $$\lim_{x\to x_0}f_a(x)^{g_a(x)}=a$$

Take $a>0$ and define $$f_a(x)=\left (1+\frac {\log(a)}{x}\right )\qquad g(x)=x$$ As you can see $f_a$ and $g$ satisfy $(1)$ in $x_0=+\infty$ for every $a$, because $$\lim_{x\to x_0}f_a(x)^{g(x)}=\lim_{x\to \infty}\left(1+\frac {\log(a)}{x} \right)^x"="1^\infty$$ Although, taking the limit, $$\lim_{x\to x_0}f_a(x)^{g(x)}=\lim_{x\to \infty}\left(1+\frac {\log(a)}{x} \right)^x=e^{\log(a)}=a$$ so the limit of an indeterminate form $1^\infty$ can be any positive real number.

3

There is a flow in your logic, you separated the limit then you did it part by part.

This is correct on it's own: $\lim_{x\to\infty} (1+\frac{1}{x})^x = e$

This is correct on it's own as well $\lim_{x\to\infty} (\frac{1}{x}) = 0$

This part is nonesensical jumoing to conclusion:

Thus,

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

because the correct version is this $\lim_{x\to\infty} (1+0)^x = 1$

Finally, $1^{\infty} = e$. Incorrect conclusion

Why is this incorrect? Can anyone explain where the math is wrong?

Math is not wrong, the logic is wrong, limit of a function is not same as function of the limit $(1+\frac{1}{x})^x$ is a function but you are applying the limit to some parts first then to the other parts, limit must be applied the whole, not part by part.

2

It's not necessarily incorrect. Notice that if we take the log of $1^\infty$, we have

$$\ln1^\infty=\infty\ln1=\infty\times0$$

And we all know very well that this indeterminate form can equal anything.

However, you reasoning is still wrong, since one may not move limits around like that.