-1

My attempt:

$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n}$

$=\left(1+\frac{1}{\infty}\right)^{\infty}$

$=\left(1+0\right)^{\infty}$

$=\left(1\right)^{\infty}$

$=1$

Yet this clearly cannot be correct, as $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n}=e$

Where was my mistake?

Blue
  • 75,673
David Ma
  • 65
  • 8
    Your mistake is pretending $\infty$ is a number using normal arithmetic – Henry Jan 23 '24 at 22:35
  • Am I forbidden from substituting $\infty$ into an equation? Why? – David Ma Jan 23 '24 at 22:37
  • 1
    @DavidMa because $\infty$ is not a number and the limits don't work that way. Do you know the definition of limit? Have you looked at some proof of $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$? – Marco Jan 23 '24 at 22:42
  • 1
    The limit form $1^\infty$ is indeterminate, so it cannot be simplified to $1$. Some limit forms can be simplified to a particular value while others cannot. If you don't consider them as limit forms, then yes, you are personally forbidden from substituting $\infty$ into an expression. – Brian Moehring Jan 23 '24 at 22:42
  • 4
    What you have really done is say $\lim\limits_{n \to \infty}\left(1+\frac1n\right)^n = \lim\limits_{n \to \infty}\lim\limits_{k \to \infty}\left(1+\frac1k\right)^n$ but it is not. The two $n$s increase together rather than separately. Note that $\lim\limits_{n \to \infty}\lim\limits_{k \to \infty}\left(1+\frac1k\right)^n=1$ while swapping the order of the limits gives $\lim\limits_{k \to \infty}\lim\limits_{n \to \infty}\left(1+\frac1k\right)^n=\infty$. – Henry Jan 23 '24 at 22:49
  • 1
    Actually more to the point is you are treating those two expressions of $n$ as though they were independent. You can't evaluate one while ignoring the other and then evaluate the other one later. $\lim (1 + \frac 1n)^n$ does not equal $(1 + \lim \frac 1n)^{\lim n}$. You have to find some way to evaluate there limits simultaneously. (No-one ever said it would be easy). – fleablood Jan 23 '24 at 23:09
  • 3
    Your mistake can be illustrated by $2=(2^{1/n})^n.$ According to your method the limit is $1,$ which is certainly wrong. – Ryszard Szwarc Jan 23 '24 at 23:53

2 Answers2

5

When $\underset{x\rightarrow a}{\lim}f(x)\ge0$ and $\underset{x\rightarrow a}{\lim}g(x)>0$ (both finite), you are allowed to use the property:

$\underset{x\rightarrow a}{\lim}f(x)^{g(x)}=\left(\underset{x\rightarrow a}{\lim}f(x)\right)^{\underset{x\rightarrow a}{\lim}g(x)}$.

It is important to clarify that this property is also true when $a=\infty$.

This property can sometimes be applied outside the mentioned hypotheses; one must carefully examine the case. However, I would like to emphasize that $1^\infty$ is also an indeterminate form.

Kevin
  • 735
  • 2
    The unspoken connection to the question seems to be "if it's not these cases, then you cannot use this property" which is false. I can't see how this is on-topic. – Brian Moehring Jan 23 '24 at 22:50
  • 1
    @BrianMoehring I agree, this is a fact not connected with the issue raised! – user Jan 24 '24 at 06:31
  • This answer while technically true is misleading. n is not finite. Too many 1st semester Calculus students would miss this and think the property can be applied. – nickalh Jan 24 '24 at 08:15
0

This limit needs to be done in a process resembling logarithmic differentiation.

Let y(n) be the function$$y=\left(1+\frac{1}{n}\right)^{n}$$ $$\ln y = \ln \left( \left( 1+\frac1n \right) ^n\right)$$ $$\ln y = n \cdot \ln \left( 1+\frac1n \right)$$ However, the limit of the right hand side, R.H.S., would be $\infty \cdot 0$ So rewrite before applying LHospital's rule. $$\lim_{n \to \infty} \ln y = \lim_{n \to \infty}\frac{\ln \left( 1+\frac1n \right)}{\frac1n}$$ Now both numerator and denominator go towards zero, enabling LHopital's rule. $$\lim_{n \to \infty} \ln y = \lim_{n \to \infty}\frac{ \left( 1+\frac1n \right)^{-1}\cdot n^{-2}}{n^{-2}}$$ Simplifying, this becomes $$\lim_{n \to \infty} \ln y = \lim_{n \to \infty} \left( 1+\frac1n \right)^{-1}$$ The right hand limit becomes $$\frac{1}{1+\frac{1}{\infty}}$$ if you'll allow the abuse of notation. $$\ln y= 1$$ and finally $$y=e$$

Gary
  • 31,845
nickalh
  • 1,263