L'Hopital and Mc'laurin is useless

Calculus Level 4

$\Large \lim_{x\rightarrow 0} \left( 1+x \right)^{\frac{1}{x^{x^{x}}}}$

Find the closed form of the limit above to 3 decimal places.

The answer is 2.718.

3 solutions

Chew-Seong Cheong
Mar 15, 2017

$\begin{aligned} L & = \lim_{x \to 0}(1+x)^{\frac 1{x^{x^x}}} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, see note.} \\ & = \exp \left(\lim_{x \to 0} \frac 1{x^{x^x}} \left(\frac {1+x}1-1 \right) \right) \\ & = \exp \left(\lim_{x \to 0} \frac x{x^{x^x}} \right) \\ & = \exp \left(\lim_{x \to 0} \frac 1{x^{\color{#3D99F6}x^x-1}} \right) & \small \color{#3D99F6} \lim_{x \to 0} x^x - 1 = 0 \\ & = \exp \left(\frac 11 \right) = e \approx \boxed{2.718} \end{aligned}$

Note: For a $1^\infty$ case (method 2) : $\small \displaystyle \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = \exp \left(\lim_{x \to a} h(x)\left(\frac {f(x)}{g(x)}-1\right) \right)$

first , we need to prove that if $\displaystyle \lim_{x\rightarrow 0} e^{f(x)}$ exists then $\displaystyle \lim_{x\rightarrow 0} e^{f(x)} = exp\left[\displaystyle \lim_{x\rightarrow 0} f\left( x \right) \right]$

second , i dont know why this true $\displaystyle \lim_{x\rightarrow 0} x^{x^{x}-1} = x^{\left(\displaystyle \lim_{x\rightarrow 0} x^{x}-1\right) } ?$ i mean what theorem,lemma or definition use to get it...thanks before.

uzumaki nagato tenshou uzumaki - 4 years, 2 months ago

First, I was using the second method of link : $\small \displaystyle \lim_{x \to a} \left(\frac {f(x)}{g(x)} \right)^{h(x)} = \exp \left(\lim_{x \to a} h(x)\left(\frac {f(x)}{g(x)}-1\right) \right)$ for solving $1^\infty$ . I can't find the proof.

Second, I was assuming $\displaystyle \lim_{x \to 0} x^{x^x-1} = \lim_{x \to 0} x^x = 1$ . Again no reference.

Chew-Seong Cheong - 4 years, 2 months ago

Used second formula to get to answer

subh mandal - 4 years, 2 months ago

When you substitute ( I suppose) $\lim_{x\to 0} [x^x -1]$ then simultaneously it converts to a $0^0$ not $x^0$ , thus describing it zero seems inappropriate. Correct me if I took it wrong..

Vishal Yadav - 4 years, 2 months ago

If we consider $\displaystyle \lim_{x \to 0} x^x = \lim_{x \to 0} \exp (x\ln x)$ . By L'Hôpital's Rule $\displaystyle \lim_{x \to 0} x\ln x = 0$ , therefore, $\displaystyle \lim_{x \to 0} x^x = e^0 = 1$

Chew-Seong Cheong - 4 years, 2 months ago

Viki Zeta
Mar 14, 2017

$\displaystyle \lim_{x\rightarrow 0} (1+x)^{\dfrac{1}{x^{x^{x}}}} \\ = \lim_{x\rightarrow \infty} (1+\dfrac{1}{x})^{\dfrac{1}{x}^{\dfrac{1}{x}^{\dfrac{1}{x}}}} \\ = \lim_{x\rightarrow \infty} (1 + \dfrac{1}{x})^{x} \\ = \mathrm{e} \\ \boxed{\approx 2.71828}$

are you sure that is a rigorous argument?

Rohith M.Athreya - 4 years, 2 months ago

This solution is wrong. All of the steps are unjustified.

Pi Han Goh - 4 years, 2 months ago

Yep I thought so

Rohith M.Athreya - 4 years, 2 months ago

$\displaystyle \lim_{x\rightarrow 0} \left( 1+x \right)^{\dfrac{1}{x^{x^{x}}}} = \lim_{x\rightarrow \infty} \left( 1+\dfrac{1}{x} \right)^{\dfrac{1}{\dfrac{1}{x}^{\dfrac{1}{x}^{\dfrac{1}{x}}}}}$ or equals $\displaystyle \lim_{x\rightarrow \infty} (1+\dfrac{1}{x})^{x^{x^{x}}}$ ?

uzumaki nagato tenshou uzumaki - 4 years, 2 months ago

EDITED. Thanks!

Viki Zeta - 4 years, 2 months ago

thats different too .

uzumaki nagato tenshou uzumaki - 4 years, 2 months ago

@Uzumaki Nagato Tenshou Uzumaki – wolfram alpha check. Did that solve your query?

Viki Zeta - 4 years, 2 months ago

Nice way of solving it.

Hana Wehbi - 4 years, 2 months ago

Arghyadeep Chatterjee
Mar 20, 2019

I do not think that x^x^x is defined for x < 0 . So the limit should be for x tending to 0+. And as per the solution, I used @Chew-Seong Cheong method.

L'Hopital and Mc'laurin is useless

The answer is 2.718.

3 solutions

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