Evaluating $\lim_{x\to 0} x^{x^{x}}-x^x$ using a graph

Question

I came across a question

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

I tried plotting the graph, but graph of $x^x$ doesn't exist but $x^{x^x}$ which is quite indigestible.

When I plotted the graph for $x^{x^x} - x^x$ then this graph exists which gives $-1$ at $x=0$, so I'm totally confused and not able to comprehend it.

Any help is welcomed.

Answer 1

First find out the limit for $$\lim_{x\to 0} x^x$$ $$x=e^{\ln x} \rightarrow \lim_{x\to 0}x^x=\lim_{x\to 0}e^{\ln x^x} = \lim_{x\to 0}e^{x\ln x}=\lim_{x\to 0}e^{\frac{\ln x}{\frac1x}}$$ Apply L'hospital for $\frac{\ln x}{\frac1x}$ $$\lim_{x\to 0}e^{x}\rightarrow1$$ Now $$\lim_{x\to 0}x^{x^x}\rightarrow e^{x^x\ln x}$$ We know $\lim_{x\to 0}x^x=1$ $$\lim_{x\to 0}e^{\ln x}=\lim_{x\to 0}x=0$$ Hence $x^{x^x}-x^x=-1$

Answer 2

Let $f(x)= x$ and $g(x)= x^x$ for all $x\in\mathbb {R^+}$ . Notice that $\displaystyle \lim_{x\to 0} g(x) $ does exist which is $1$ and non-zero so $$\lim_{x\to 0}(f(x))^{g(x)}-g(x)) =\lim_{x\to 0} f(x)^{g(x)} - \lim_{x\to 0} g(x) = \lim_{x\to 0} (f(x))^1-1= 0-1=-1$$

Answer 3

Here is the graph for $x^x$: Graph of $x^x$

And the graph of $x^{x^x}$

Finally, the graph of $x^{x^x} -x^x$

Yes, you're correct that at $x=0$ we have the value of graph as $-1$, but please notice that the graph isn't defined for negative $x$, it is because $x^x$ is imaginary if $x$ is negative and not an integer. For example, let's take $x= -1/4$, then we have $$ x^x = (-1/4)^{-1/4} \\ x^x = \frac{1}{ \left( \left( \frac{-1}{4}\right )^{1/2}\right)^{1/2}} $$ Clearly, the square root (that is 1/2 power) of a negative number is imaginary.

So, we don't have left hand limit defined for $\lim_{x \to 0} x^{x^x} -x^x$ and hence the limit doesn't exist. The correct way to write our observation is $$ \lim_{x \to 0^{+}} x^{x^x} -x^x = -1 $$

Hope it helps in some way!

Answer 4

First of all, to avoid complex numbers, we should look at $\lim_{x\to 0^+}$.

We are going to use two things, $a^b = e^{b\ln a}$ and $\lim_{x\to a} e^{f(x)} = e^{\lim_{x\to a}f(x)}$.

Thus, $$\lim_{x\to 0^+} x^x = \lim_{x\to 0^+}e^{x\ln x} = e^{\lim_{x\to 0^+}(x\ln x)}.\tag{1}$$

Now, $\lim_{x\to 0^+}(x\ln x)$ is of the form $0\cdot (-\infty)$, so special care is needed. We could write it in the form $\frac{\ln x}{1/x}$ and use l'Hospital's rule to conclude that the limit is $0$. Returning to $(1)$ we get

$$\lim_{x\to 0^+}x^x = e^0 = 1.\tag{2}$$

So, $$\lim_{x\to 0^+}x^{x^x} = \lim_{x\to 0^+}e^{x^x \ln x} = e^{\lim_{x\to 0^+}(x^x \ln x)} = \left[e^{1\cdot(-\infty)} = e^{-\infty}\right] = 0.\tag{3}$$

Combining $(2)$ and $(3)$ gives us $$\lim_{x\to 0^+}(x^{x^x}-x^x) = 0 - 1 = -1.$$

Answer 5

Since you already received good answers and that you play with the graph, let me continue further that the limit itself.

Composing Taylor series $$x^x=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+\frac{1}{6} x^3 \log ^3(x)+O\left(x^4\right)$$ $$x^{x^x}=x+x^2 \log ^2(x)+\frac{1}{2} x^3 \left(\log ^4(x)+\log ^3(x)\right)+O\left(x^4\right)$$ $$x^{x^x}-x^x=-1+x (1-\log (x))+\frac{1}{2} x^2 \log ^2(x)+\frac{1}{6} x^3 \log ^3(x) (3 \log (x)+2)+O\left(x^4\right)$$

Plot the lhs and the rhs on the same graph. They almost overlap for $0 \leq x \leq 1$ (the maximum error is $0.0035$)