What will be the value of $\frac{1-x^x}{x \log x}$ as $x \rightarrow 0$? We were taught L'hopital's rule and this is one of the questions from the assignment. I can't wrap my head around the indeterminate form that we get when we put in the limit.
By equivalence.


$$1-x^x=1-e^{x\ln (x)}\sim -x\ln (x) $$
since $$\lim_{0^+}x\ln (x)=0$$
thus your limit is $-1$.


  By usual limit.


Put $t=x\ln (x) $.

as $x\to 0^+ \;\; , t \to 0$.

the limit is
$$\lim_{t\to 0}\frac {-(e^t-1)}{t}=-1$$


  By l'Hospital.


Put $f (x)=x\ln (x) $.

$$\lim_{0^+}\frac {1-e^{f (x)}}{f (x)}=$$
$$\lim_{0^+}\frac {-f'(x)e^{f (x)}}{f'(x)}=$$
$$\lim_{0^+}(- e^{f (x)})=-e^0=-1$$
Note:
$$\lim_{x\to 0} x^x=1.$$
Hence:
$$\lim_{x\to 0} \frac{1-x^x}{\ln x^x}=\lim_{y\to 1} \frac{1-y}{\ln y}=LR=\lim_{y\to 1} \frac{-1}{\frac{1}{y}}=-1.$$