Solve $\lim_{x\to 0^{+}}x^{x^x-1}$

Question

$$\lim_{x\to 0^{+}}x^{x^x-1}$$

Note:- I have solved this problem and below is the solution but I am searching for the better approach

My attempt is as follows:-

$x^x$ is indeterminate form $(0^0)$ as $x$ tends to $0^{+}$

$$\lim_{x\to 0^{+}}x^{x^x-1}=e^{\lim_{x\to 0^{+}}\left(x^x-1\right)ln(x)}\tag{1}$$

Let's assume $\lim_{x\to 0^{+}}\left(x^x-1\right)ln(x)=y$

$$y=\lim_{x\to 0^{+}}\left(e^{x\ln x}-1\right)\ln x\tag{2}$$

As $x$ tends to $0^{+}$, $x\ln x$ is the indeterminate form $(0\cdot\infty)$

So first let's see what this indeterminate form actually tends to

$$z=\lim_{x\to 0^{+}}x\ln x$$

Assume $t=\ln x$

$$z=\lim_{t\to -\infty}e^tt$$ $$z=\lim_{t\to -\infty}\dfrac{t}{e^{-t}}$$

So we are getting $\dfrac{-\infty}{\infty}$ and we can apply L's Hospital rule here

$$z=\lim_{t\to -\infty}\dfrac{1}{-e^{-t}}$$

$$z=0$$

So going back to equation $(1)$, now we can say as $x$ tends to $0^{+}$, $x\ln x$ tends to $0$

$$y=\lim_{x\to 0^{+}}\left(e^{x\ln x}-1\right)\ln x\tag{3}$$

$$y=\lim_{x\to 0^{+}}\dfrac{\left(e^{x\ln x}-1\right)}{x\ln x}\cdot x\ln^2 x$$

$$y=\lim_{x\to 0^{+}}\dfrac{\left(e^{x\ln x}-1\right)}{x\ln x}\cdot x\ln^2x$$

$$y=\lim_{x\to 0^{+}}x\ln^2x$$

$$y=\lim_{x\to 0^{+}}\dfrac{ln^2x}{\dfrac{1}{x}}$$

So we have $\dfrac{\infty}{\infty}$ form here and so we can apply L's hospital rule here

$$y=\lim_{x\to 0^{+}}\dfrac{2\ln x\cdot\dfrac{1}{x}}{\dfrac{-1}{x^2}}$$

$$y=\lim_{x\to 0^{+}}\dfrac{2\ln x}{\dfrac{-1}{x}}$$

Again applying L's hospital rule here as we have $\dfrac{\infty}{\infty}$ form

$$y=\lim_{x\to 0^{+}}\dfrac{\dfrac{2}{x}}{\dfrac{1}{x^2}}$$ $$y=\lim_{x\to 0^{+}}2x$$ $$y=0$$

Putting value of $y$ back in equation $(1)$

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

So lots of ups and down in this problem, can it solved by any simpler means?

Answer 1

$$x^{x^x-1}=e^{(x^x-1)\ln{x}}=e^{\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot x\ln^2{x}}=e^{\frac{e^{x\ln{x}}-1}{x\ln{x}}\cdot4\left(\sqrt{x}\ln\sqrt{x}\right)^2}\rightarrow e^0=1.$$

Answer 2

We have $$\lim_{x\to0^+}x^{x^x-1}=\lim_{x\to0^+}\frac{x^{x^x}}{x}\overset{\text{L'H}}{=}\lim_{x\to0^+}\frac{d}{dx}(x^{x^x}).$$

Can you take it from here?

Answer 3

Another use of LH: since $\frac{x^x-1}{1/\ln x}\sim\frac{x^x(\ln x+1)}{-1/(x\ln^2x)}=-\underbrace{x\ln^3 x}_{\to0}\underbrace{\left(1+\frac{1}{\ln x}\right)}_{\to1}\underbrace{x^x}_{\to1}\to0$, your limit is $1$.

Answer 4

$\lim_{x\to0^+}x^{x^x-1}=\lim_{x\to0^+}\frac{x^{x^x}}{x} = \lim_{x\to0^+}\frac{x^{e^{xlog(x)}}}{x} $

Now, $\lim_{x\to0^+} xlog(x) = \lim_{x\to0^+}\frac{log(x)}{\frac{1}{x}}\overset{\text{L'H}}{=} \lim_{x\to0^+}\frac{\frac{1}{x}}{\frac{-1}{x^2}} = 0$

When we plug that back in, we get

$\lim_{x\to0^+}\frac{x^{x^x}}{x} = \lim_{x\to0^+}\frac{x^{e^{0}}}{x} = \lim_{x\to0^+}\frac{x^1}{x} = 1$