solving [MATH]\displaystyle\lim_{x \to \infty}\bigg(1+\frac{5}{x}\bigg)^x[/MATH]

[burt]

[MATH]\displaystyle\lim_{x \to \infty}\bigg(1+\frac{5}{x}\bigg)^x[/MATH] I was given this problem. The answer key gives the answer as [MATH]e^x[/MATH]. I want to know why the answer isn't one. When x approaches infinity, [MATH]\frac5x[/MATH] becomes zero. [MATH](1+0)^{\infty}=1[/MATH]

 

[Mr. Bland]

You sure it says [MATH]e^x[/MATH]? The limit as stated is something different.

While it's true that [MATH]\displaystyle\lim_{x \to \infty}\frac{5}{x} = 0[/MATH], the operation [MATH]x^\infty[/MATH] is undefined. Bridging the gap on paper may feel arbitrary, but when in doubt, you can plot a graph experimentally of the original expression and find that it does indeed converge.

 

[burt]

gives the answer as exex\displaystyle e^x.

You sure it says exex\displaystyle e^x? The limit as stated is something different.

Thank you - it says \[e^5\]

 

[burt]

he operation x∞x∞\displaystyle x^\infty is undefined.

Why is \(1^{\infty}\) undefined? one to any power is one.

 

[MarkFL]

I would begin by stating:

[MATH]L=\lim_{x\to\infty}\left(\left(1+\frac{5}{x}\right)^x\right)[/MATH]
Take the natural log of both sides:

[MATH]\ln(L)=\ln\left(\lim_{x\to\infty}\left(\left(1+\frac{5}{x}\right)^x\right)\right)[/MATH]
[MATH]\ln(L)=\lim_{x\to\infty}\left(\ln\left(\left(1+\dfrac{5}{x}\right)^x\right)\right)[/MATH]
[MATH]\ln(L)=\lim_{x\to\infty}\left(\frac{\ln\left(1+\dfrac{5}{x}\right)}{\dfrac{1}{x}}\right)[/MATH]
Now we have the indeterminate form 0/0...can you apply L'Hôpital's Rule?

 

[Mr. Bland]

Why is \(1^{\infty}\) undefined? one to any power is one.

Infinity doesn't behave like a number, and consequently can't be used in most numerical contexts. It is, by nature, indeterminate, and treating it like a number can lead to contradictions. For instance, given that [MATH]\infty + 1 = \infty[/MATH], you could reason that if variable [MATH]x = \infty[/MATH], then [MATH]x + 1 = x[/MATH] (a clear contradiction).

Further reading: https://en.wikipedia.org/wiki/Indeterminate_form

 

[burt]

Infinity doesn't behave like a number, and consequently can't be used in most numerical contexts. It is, by nature, indeterminate, and treating it like a number can lead to contradictions. For instance, given that [MATH]\infty + 1 = \infty[/MATH], you could reason that if variable [MATH]x = \infty[/MATH], then [MATH]x + 1 = x[/MATH] (a clear contradiction).

Further reading: https://en.wikipedia.org/wiki/Indeterminate_form

True, but \(1^{\infty}\) is not like that. one to any number, no matter what is always 1. So, why isn't that the answer?

 

[burt]

I still don't understand why my way doesn't work. \(\lim_{x\to\infty}1^x=1\)?

 

[pka]

I still don't understand why my way doesn't work. \(\lim_{x\to\infty}1^x=1\)?

Burt, It hard to tell about the strength of your calculus background. In a usual Calculus II course students meet Napper inequality.
It states that if \(\displaystyle 0<a<b\) then \(\displaystyle \dfrac{1}{b}<\dfrac{\log(b)-\log(a)}{b-a}<\dfrac{1}{a}\) proved with the mean value theorem.
With that inequality we prove that \(\displaystyle \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = e\)
Moreover if \(\displaystyle a\cdot b\cdot c \ne 0\) then \(\displaystyle \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{b}{x+a}} \right)^{cx}} = e^{bc}\)

 

[Cubist]

I still don't understand why my way doesn't work. \(\lim_{x\to\infty}1^x=1\)?

You seem to be thinking that the limit can be considered as two separate limits which can be evaluated individually, one at a time, like this...
[MATH]\lim_{y \to \infty} \left( \lim_{x \to \infty}\left(1+\frac{5}{x}\right) \right)^y[/MATH]
...but in the original question the 5/x within the bracket interacts with the "^x" outside the bracket as x grows
[MATH]\lim_{x \to \infty}\left(1+\frac{5}{x}\right)^x[/MATH]

 

[Cubist]

Moreover if \(\displaystyle a\cdot b\cdot c \ne 0\) then \(\displaystyle \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{b}{x+a}} \right)^{cx}} = e^{bc}\)

I think \(\displaystyle a\) can be zero, but \(\displaystyle b\cdot c \ne 0\)

 

[pka]

I think \(\displaystyle a\) can be zero, but \(\displaystyle b\cdot c \ne 0\)

Yes thank you that is correct. In fact in most cases we want a=0.
I keep telling myself do not cut and paste.