Let L=\displaystyle \lim_{x\to0}\left(1+ax^{k}+ bx^{k+1}+ cx^{k+2}\right)^{x^{-k+1}} .
Then L=\displaystyle \lim_{x\to0}\exp \left(\ln\left[\left(1+ax^{k}+ bx^{k+1}+ cx^{k+2}\right)^{x^{-k+1}}\right]\right) ,
     because \exp and \ln are inverses.

It is a property of logarithms that \ln\left[A^B\right]=[B][\ln A].
     Use A=1+ax^{k}+ bx^{k+1}+ cx^{k+2}    and     B=x^{-k}   .

We get L=\displaystyle \lim_{x\to0}\exp \left(\left[x^{-k+1}\right]\left[\ln\left(1+ax^{k}+ bx^{k+1}+ cx^{k+2}\right)\right]\right) .

Next, use that \lim and \exp commute.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-k+1}\right]\left[\ln\left(1+ax^{k}+ bx^{k+1}+ cx^{k+2}\right)\right]\right) .

Fact:     \ln\left(1+[f(x)]\right)\quad\sim\quad f(x)\quad\hbox{as }x\to0,
         provided      f(x)\to0\quad\hbox{as }x\to0.
Use this fact, with f(x)=ax^{k}+ bx^{k+1}+ cx^{k+2}.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-k+1}\right]\left[ax^{k}+ bx^{k+1}+ cx^{k+2}\right]\right) .

A polynomial in x is asymptotic to its lowest order term, as x\to0.
Then      ax^{k}+ bx^{k+1}+ cx^{k+2}\quad\sim\quadax^{k},\quad \hbox{as }x\to0.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-k+1}\right]\left[ax^{k}\right]\right) .

Then L=\displaystyle \exp\left(\lim_{x\to0}\,\,ax\right) ,   because    \left[x^{-k+1}\right]\left[ax^{k}\right]=ax \hbox{ on }x\ne0.

Then L=\displaystyle \exp\left(0\right)=1 ,      because \displaystyle\lim_{x\to0}\,\,ax=0.

Now enter 0 into the Answer box, to indicate that the answer is \exp(0/2).