Puny Powers

Calculus Level 1

True or false :

$\LARGE \lim_{x\to0^+} x = \lim_{x\to0^+} x^{x^x}$

True False

2 solutions

Jesse Nieminen
May 29, 2016

Relevant wiki: Limits by Logarithms

Let's first prove the following lemma.

$\text{Claim: }$

$\large\displaystyle\lim_{x \to 0^+} x^x = 1$

$\text{Proof: }$

Clearly, $\large\displaystyle\lim_{x \to 0^+} x \ln x = - \displaystyle\lim_{x \to 0^+} \frac{ \ln x}{-x^{-1}}$ .

Since, $\large\displaystyle\lim_{x \to 0^+} \ln x = \displaystyle\lim_{x \to 0^+} -x^{-1} = -\infty$ , L'Hôpital's Rule can used.

Now, $\large- \displaystyle\lim_{x \to 0^+} \frac{ -\ln x}{x^{-1}} = - \displaystyle\lim_{x \to 0^+} \frac{\frac{1}{x}}{x^{-2}} = - \displaystyle\lim_{x \to 0^+} x = 0$ .

Thus, $\large \displaystyle\lim_{x \to 0^+} x \ln x = 0 \Rightarrow \displaystyle\lim_{x \to 0^+} e^{x \ln x} = \displaystyle\lim_{x \to 0^+} x^x = 1$ . $\boxed{}$

Using the lemma we find that $\large \displaystyle \lim_{x\to0^+} x^x \ln x = \displaystyle \lim_{x\to0^+} \ln x = -\infty \Rightarrow \displaystyle \lim_{x\to0^+} x^{x^x} = \displaystyle \lim_{x\to0^+} e^{x^x \ln x} = 0 = \displaystyle \lim_{x\to0^+} x$ . $\boxed{ }$

Thus, the answer is $\boxed{\text{True}}$ .

You're almost correct

Using the lemma we find that $\large \lim_{x \to 0^+} x^{x^x} = \left(\lim_{x\to0^+} x\right)^\left( \lim_{x \to 0^+} x^x\right) = \lim_{x \to 0^+} x$ . $\boxed{ }$

This line is wrong. Using your logic, we should get $\large \displaystyle \lim_{x\to0^+} x^x =\left(\lim_{x\to0^+} x \right)^{\displaystyle \left(\lim_{x\to0^+} x \right)} = 0^0$ .

Pi Han Goh - 5 years ago

Is it satisfactory to state that $\large \displaystyle \lim_{x\to c} f\left(x\right)^{\displaystyle g\left(x\right)} = \displaystyle\left(\lim_{x\to c} f\left(x\right)\right)^{\displaystyle \left(\lim_{x\to c} g\left(x\right)\right)}$

if $\large \displaystyle \lim_{x\to c} f\left(x\right) \geq 0$ and $\large \displaystyle \lim_{x\to c} f\left(x\right) \neq 0$ if $\large \displaystyle\lim_{x\to c} g\left(x\right) = 0$ , or is this even a valid statement?

Jesse Nieminen - 5 years ago

Unfortunately, this is not correct. Counterexample: $f(x) = 1 + \dfrac1x$ and $g(x) = x$ , set $c=0$ .

Pi Han Goh - 5 years ago

@Pi Han Goh – What if we assume $\large \displaystyle \lim_{x\to c} f\left(x\right)$ to be finite?

Jesse Nieminen - 5 years ago

@Jesse Nieminen – Hmmm, I don't know where this conversation is heading to. I've already shown you that it doesn't work for $f(x) = g(x) = x$ .

If you follow your argument as stated below:

if $\large \displaystyle \lim_{x\to c} f\left(x\right) \geq 0$ and $\large \displaystyle \lim_{x\to c} f\left(x\right) \neq 0$ if $\large \displaystyle\lim_{x\to c} g\left(x\right) = 0$ , or is this even a valid statement?

Then your $f(x)$ does not satisfy this condition.

Pi Han Goh - 5 years ago

@Pi Han Goh – Oh I see that you have edited your solution. It's correct now! +1

Pi Han Goh - 5 years ago

@Pi Han Goh – I mean, is $\displaystyle \lim_{x\to c} f(x)^{\displaystyle g(x)} = \displaystyle \lim_{x\to c} f(x) ^ {\displaystyle \lim_{x\to c} g(x)}$ valid if:

$\displaystyle \lim_{x\to c} f(x) \geq 0$

$\displaystyle \lim_{x\to c} f(x) \neq \pm \infty$

$\displaystyle \lim_{x\to c} f(x) \neq 0$ if $\displaystyle \lim_{x\to c} g(x) = 0$ or if $\displaystyle \lim_{x\to c} g(x) = \pm \infty$

Jesse Nieminen - 5 years ago

@Jesse Nieminen – Still incorrect. Set $c$ as $\infty$ , and $f(x) = x^{1/x} , g(x) = x$ .

You might be interested to read up indeterminate forms .

Pi Han Goh - 5 years ago

@Pi Han Goh – So is $\displaystyle \lim_{x\to c} f(x)^{\displaystyle g(x)} = \displaystyle \lim_{x\to c} f(x) ^ {\displaystyle \lim_{x\to c} g(x)}$ valid if we assume that $f(c)^{\displaystyle g(c)}$ is not of an indeterminate form?

Jesse Nieminen - 5 years ago

@Jesse Nieminen – Correct! =D

Pi Han Goh - 5 years ago

@Pi Han Goh – nice counterexample

Mehdi K. - 5 years ago

I have a query: LHS is a positive number very close to zero. On RHS, the term $x^{x^x}$ is actually very small positive number twice raised to a power which is positive but <1. Now we know that raising a proper fraction to the power of another proper fraction increases the value. Therefore RHS>LHS.

I would be very thankful if somebody would tell me where I've gone wrong.

Thanks in advance! :-)

Aniruddha Bhattacharjee - 4 years, 12 months ago

Now we know that raising a proper fraction to the power of another proper fraction increases the value. Therefore RHS>LHS.

Not true. Your logic is incorrect.

Pi Han Goh - 4 years, 12 months ago

I know that it is incorrect. I am curious to know where it is wrong.

Thanks

Regards

Aniruddha Bhattacharjee - 4 years, 12 months ago

@Aniruddha Bhattacharjee –

Therefore RHS>LHS.

Why must this line be true? You made no real argument to justify that this is true. You only showed that it "increases the value", but that has no relation to LHS at all.

Pi Han Goh - 4 years, 12 months ago

@Pi Han Goh – I said that the term on the RHS is x^x i.e. a proper fraction raised to a proper fraction. Therefore a proper fraction (in this case, 'x'), when raised to a proper fraction (in this case, 'x'), yields a value higher in magnitude than the value used as base in the exponent calculation (in this case, just 'x', as 'x' is used as base in 'x^x'). Therefore RHS>LHS.

I hope I'm clearer now. Regards

Aniruddha Bhattacharjee - 4 years, 12 months ago

@Aniruddha Bhattacharjee – I understood what you said, I'm saying that your argument is not sound. By your logic,

$\Large \lim_{x\to0^+} x^{x^x} < \lim_{x\to0^+} x^{x^{x^{x^x}}}$

is true, when in fact it's false.

Pi Han Goh - 4 years, 12 months ago

@Pi Han Goh – I know it is false. I'm asking where the discrepancy stands in my argument. Regards

Aniruddha Bhattacharjee - 4 years, 12 months ago

@Aniruddha Bhattacharjee – I've given you a counterexample to your argument already, you should know from my counterexample that your argument is wrong. Or are you saying that you don't understand my counterexample?

Pi Han Goh - 4 years, 12 months ago

The RHS is of course 1 but the LHS is 0. The answer should be false.

A Former Brilliant Member - 4 years, 11 months ago

No. Both are equal to 0.

Pi Han Goh - 4 years, 11 months ago

Lance Fernando
Jul 22, 2016

It's just pretty simple. We all know that limit as x approaches to 0 from the right in the function of x is 0. Proving that the same thing happens in x raised to x raised to x, take a look at these values: 1^1^1 = 1 0.5^0.5^0.5 <1 0.25^0.25^0.25 <<1 0.1^0.1^0.1<<<1 It approaches therefore to 0.

Puny Powers

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