Infinite Product Of 1's

Calculus Level 1

$\begin{aligned} \lim_{x\to0}(1+x)^{1/x}&=&(1+0)^{1/0} \\ &=&1^\infty\\ &=&1\times1\times1\times\cdots\times1 \\ &=& 1 . \end{aligned}$

Is the working above correct?

Yes, it is correct No, it is incorrect

4 solutions

A Former Brilliant Member
May 29, 2016

$1^{\infty}$ is an indeterminate form and it does not equal 1.

The better way of phrasing your solution is

$1 ^ \infty$ is an indeterminate form and the limit need not equal 1. In fact, we can show that it is not 1 because ...

Not all $1 ^ \infty$ indeterminate forms are not 1.

Calvin Lin Staff - 5 years ago

Then, is $\displaystyle \lim_{n\to\infty} 1^n$ an indeterminate form as well? Why or why not?

Pi Han Goh - 5 years ago

$1^{\infty}$ is an intermediate form because we don't know if it is $0.9999999...9^{\infty} =0$ or $1.0000000...1^{\infty} = \infty$ , but in your example we know that the base of the exponent is exactly equal to $1$ , so $1^{\infty}=1$

Sabhrant Sachan - 5 years ago

This is not an an In determinant form as we have exact 1 multiplying infinite times which is exactly One $1$ .

Rishabh Deep Singh - 5 years ago

No. That is 1. You can't use the substitution method for limits that approach with infinity.

A Former Brilliant Member - 5 years ago

Matteo Monzali
May 29, 2016

I think that if we put y=1/x the limit becomes: $\lim _{ x\rightarrow 0 }{ (1+x)^{ 1/x } } =\lim _{ y\rightarrow \infty }{ (1+\frac { 1 }{ y } )^{ y } } = e$

Wouldn't we (still) get

$\lim_{y\to\infty} \left(1 + \dfrac1y\right)^y = \left(1 + \dfrac1\infty\right)^\infty = 1^\infty = 1\times1\times1\times\cdots\times1 = 1?$

Pi Han Goh - 5 years ago

I can prove that $\lim_{y\to\infty} \left( 1 + \dfrac1y \right)^y = e$ and $\lim_{x\to 0} (1 + x)^{\frac{1}{x}} = e$ . The problem with this last limit is that you also have to consider lateral limits, althought it can also to be avoided if we use logarhitms and McLaurin series... More generally, if $f$ is a function such that $\lim_{x\to a} f(x) = 0$ then $\lim_{x \to a} ( 1 + f(x) )^{\frac{1}{f(x)}} = e$

Guillermo Templado - 5 years ago

No, in fact there are some limits that are determinated (and that you can't calculate), such as

$\lim _{ x\rightarrow \infty }{ (1+\frac { 1 }{ x } )^{ x } } = e$

$\lim _{ x\rightarrow 0 }{\frac { 1 -\cos x}{ x^2 }} =\frac{1}{2}$

For more info see wikipedia

Matteo Monzali - 5 years ago

They can be "calculated".

Pi Han Goh - 5 years ago

Viki Zeta
Jun 6, 2016

We have, $(1 + 0)^{\frac{1}{0}} = \sqrt[0]{(1 + 0)}$ , which is undefined. So the proof is not correct.

Patience Patience
Jun 6, 2016

u put here(1+x) x= 0 , then u put 1\x x= 0 1\0 = - or + infinty rong

Infinite Product Of 1's

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