Implicit Functions

Calculus Level 5

If $f(x)$ is a function implicitly defined such that ${f(x)}^{{f(x)}^{f(x)^{f(x)^{\dots}}}}=x$ , then what is $\displaystyle \ln\left( \lim_{x \to 0}\left(\frac{ f(x+1)}{\log_{x+1} f(f(x+1))}\right)^\frac{1}{x} \right)$ ?

The answer is 1.

1 solution

Albert Yiyi
Oct 21, 2018

$\begin{aligned} f(x)^x &= x \\ f(x) &= x^\frac{1}{x} \\ f(x+1) &= (x+1)^{\frac{1}{x+1}} \\ f(f(x+1)) &= (x+1) \wedge (x + 1) ^{ -\frac{1}{x + 1} - 1} \\ \log_{x+1} f(f(x+1)) &= (x + 1) ^{ -\frac{1}{x + 1} - 1} \\ \frac{f(x+1)}{\log_{x+1} f(f(x+1))} &= (x + 1) ^ \frac{x+3}{x+1} \\ \ln \lim (...)^{\frac{1}{x}} &= \ln \lim (1+x)^{\frac{1}{x} \times \frac{x+3}{x+1}} \\ &= \ln e^\frac{0+3}{0+1} \\ &= 3 \end{aligned}$

edit: thanks to Parth and Anon for pointing out the mistake, the answer should be 3, not 1.

Yes that is correct but the required limit, according to me, should be $e^3$ . The answer must be 3

Parth Sankhe - 2 years, 7 months ago

i've added more steps, hope it helps.

albert yiyi - 2 years, 7 months ago

i've made a mistake, u are right, the answer is 3.

albert yiyi - 2 years, 7 months ago

There is an error on line 4: if f(x+1)=(x+1)^(1/(x+1)), then f(f(x+1))=f(x+1)^(1/f(x+1))=((x+1)^(1/(x+1)))^(1/((x+1)^(1/(x+1)))), not what is written on line 4. I believe the question is wrong, and that the solution is 3. This is shown graphically here: https://www.desmos.com/calculator/ht0zxmyzxc

A Former Brilliant Member - 2 years, 7 months ago

i've made a mistake, u are right, the answer is 3.

albert yiyi - 2 years, 7 months ago

Implicit Functions

The answer is 1.

1 solution

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