Well, I'm doing this and I already find a resolution. I can write that limit as: $e^{\lim\limits_{n\to\infty}\frac{1}{n}}$ and that results in $1$.

My question is: Why can I do this?

  • $\begingroup$ If You know that e^x is continuous then you can do this. And the result is correct. $\endgroup$ Sep 28 '15 at 23:18
  • $\begingroup$ yes, but why can i do that? and why it have to be continuous? $\endgroup$
    – Chittolina
    Sep 28 '15 at 23:23
  • $\begingroup$ Continuity is the property that lim f(x)=f(lim x). $\endgroup$ Sep 28 '15 at 23:32

A function $f(x)$, is continuous at a point, $a$ if:

$$\lim\limits_{x\to a} f(x) = f(a)$$

This is the definition of continuity. So if $n\to \infty$, then $\frac1n \to 0$, so:

$$\lim\limits_{n\to\infty} e^{1/n}=\lim\limits_{x\to 0} e^x=e^0=1$$

because $e^x$ is continuous at $0$.


You can do it because when $n$ gets really big $ \frac{1}{n}$ is going to get really close to zero, and anything to the zeroth power is always one. $e^{0} = 1$. I'm not sure if that answers your question though.

Jeez, constant downvotes. Anyone care to tell me what I did wrong?

  • $\begingroup$ Seriously, what's different about this answer compared to the top answer other than the fact that they stated the definition of continuity? $\endgroup$ Sep 29 '15 at 0:14
  • $\begingroup$ Because by your logic, $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$ should equal 1. Because "when $n$ gets really big, $1/n$ is going to get really close to 0" so the limit is $1^\infty$ which is clearly 1. "Stated the definition of continuity" is exactly the answer to this problem, not some "really big" "really small" arguments. The point is you can't make these hand-waving heuristic arguments as an answer. Maybe they provide intuition for basic limits, but for trickier problems, that is rarely the case. $\endgroup$
    – user217285
    Sep 29 '15 at 0:16
  • $\begingroup$ Yeah, you're right... it was kind of a hand-waving argument. I figured the "really big" argument was an okay thing to do since there was only one instance of x in the expression, so in theory I wouldn't have to worry about how other parts of the expression change as x grows large, but that's pretty lazy and really just plain wrong. Thanks for the clarification. $\endgroup$ Sep 29 '15 at 0:37
  • $\begingroup$ @Nitin I actually have one more quick question: Why does the domain of f(x) = x change when you multiply f(x) = x by x/x? Shouldn't it be the same function algebraically? Apparently not. What's going on here? $\endgroup$ Sep 29 '15 at 0:57
  • $\begingroup$ because $f(x) = x$ and $g(x) = \frac{x^2}{x}$ aren't the same functions. They are when $x \neq 0$, but the latter isn't defined when $x = 0$ ($g(0) = \frac{0}{0}$ which is meaningless). $\endgroup$
    – user217285
    Sep 29 '15 at 1:20

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