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. 
 A: 
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) $.
$$\lim_{0^+}\frac {1-e^{f (x)}}{f (x)}=$$
$$\lim_{0^+}\frac {-f'(x)e^{f (x)}}{f'(x)}=$$
$$\lim_{0^+}(- e^{f (x)})=-e^0=-1$$
A: Since $\lim\limits_{x\to0^+}x\log(x)=0$, we get
$$
\begin{align}
\lim_{x\to0^+}\frac{1-x^x}{x\log(x)}
&=\lim_{x\to0^+}\frac{1-e^{x\log(x)}}{x\log(x)}\\
&=\lim_{u\to0}\frac{1-e^u}{u}\\
&=\lim_{u\to0}\frac{-e^u}{1}\\[6pt]
&=-1
\end{align}
$$
A: as $\lim _{ x\rightarrow 0 }{ x\log { x } =0 } $$$\lim _{ x\rightarrow 0 }{ \frac { 1-{ x }^{ x } }{ x\log { x }  }  } =\lim _{ x\rightarrow 0 }{ \frac { 1-{ e }^{ x\log { x }  } }{ x\log { x }  }  } \overset { L'Hospital }{ = } \ =\lim _{ x\rightarrow 0 }{ \frac { -{ x }^{ x }\left( \log { x+1 }  \right)  }{ \left( \log { x+1 }  \right)  }  } =-\lim _{ x\rightarrow 0 }{ { e }^{ x\log { x }  } } =-1$$
