4

Evaluate $$\lim_{x \to 0^+ } x^{x^{x}} - x^x$$

This is a solved example in my text book but i do not think that the solution is quite correct.

They have essentially used the fact $\lim_{x\to0^+}x^x$ is 1 and used that to write the term to be evaluated as $$0^1 - 1$$ which gives an answer of -1 The graph indeed gives the limit at $0^+$ as -1.

BUT

We could have used $\lim_{x\to0^+}x^x$ to evaluate $\lim_{x \to 0^+ } x^{x^{x}} - x^x$ as $$1^0 - 1$$ which does not give the correct answer.

Is the book's method correct?

Avyansh Katiyar
  • 364

2 Answers2

5

Note that you solve for the exponent first.

$$a^{b^c} \color{red}{\neq (a^b)^c = a^{bc}}$$

For example,

$$2^{3^2} = 2^9 = 512 \color{red}{\neq (2^3)^2 = 2^6 = 64}$$

Therefore, when looking at $x^{x^x}$, you have to start from the exponent $x^x$ and work your way down rather than the other way around.

KM101
  • 7,176
3

The given solution uses that $x^x \to 1$ and

$$(x)^{(x^{x})} - x^x \to 0^1-1=-1$$

while you are considering

$$(x^x)^{x} - x^x =x^{(x^2)}-x^x\to 1-1=0$$

indeed

  • $(x)^{(x^{x})}=e^{x^x\log x} \to 0$ (since $x^x \log x \to -\infty$)

but

  • $x^{(x^2)}=e^{x^2\log x} \to e^0=1$
user
  • 154,566