CMI UG Power

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L &= \lim_{x \to 0^+} \left(x^{x^x}- x^x\right) \\ &= \lim_{x \to 0^+} \exp (\ln x^{x^x}) - \lim_{x \to 0^+}\exp(\ln x^x) \\ &= \exp \left(\lim_{x \to 0^+}{\color{#3D99F6}x^x}\ln x\right) - \exp \left({\color{#3D99F6}\lim_{x \to 0^+}x\ln x}\right) & \small \color{#3D99F6} \text{Note that } \lim_{x \to 0} x \ln x = 0 \implies \lim_{x \to 0} x^x = 1 \text{ (see note)} \\ & = e^{\ln 0} - e^0 \\ & = 0 - 1 \\ & = \boxed{-1} \end{aligned}$

Note:

$\begin{aligned} \lim_{x \to 0} x \ln x & = \lim_{x \to 0} \frac {\ln x}{\frac 1x} & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\frac 1x}{-\frac 1{x^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }x \\ & = \lim_{x \to 0} - \frac 1x \times x^2 \\ & = 0 \end{aligned}$

Take x^x as t.So now the problem stands out to be x^t-t....well t tends to 1 and x tends to 0 so the answer...😊😊

$\displaystyle L = \lim_{x\to0}x^x\quad \ln(L) = \lim_{x\to0}x\ln(x)=\,\,\,\text{(Using L'Hopital's Rule)} \lim_{x\to0}\frac{\frac1{x}}{-\frac1{x^2}} = 0$

$L = e^0 = 1$

$\lim_{x\to0}x^1-1 = \boxed{-1}$