Let L=\displaystyle \lim_{x\to0}\left(e^x+ax+ bx^2+ cx^3\right)^{1/x} .
Then L=\displaystyle \lim_{x\to0}\exp \left(\ln\left[\left(e^x+ax+ bx^2+ cx^3\right)^{1/x}\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=e^x+ax+ bx^2+ cx^3    and     B=1/x   .

We get L=\displaystyle \lim_{x\to0}\exp \left(\left[1/x\right]\left[\ln\left(e^x+ax+ bx^2+ cx^3\right)\right]\right) .

Next, use that \lim and \exp commute.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[1/x\right]\left[\ln\left(e^x+ax+ bx^2+ cx^3\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)=-1+e^x+ax+ bx^2+ cx^3.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[1/x\right]\left[-1+e^x+ax+ bx^2+ cx^3\right]\right) .

Because
    e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots
and because
    a power series is asymptotic (as x\to0) to its lowest order term,
it follows that
    -1+e^x+ax+ bx^2+ cx^3\quad\sim\quada+1x,\quad \hbox{as }x\to0.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[1/x\right]\left[a+1x\right]\right) .

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

Then L=\displaystyle \exp\left(a+1\right) ,      because a limit of a constant is the constant.

Now enter 2*(a+1) into the Answer box, to indicate that the answer is \exp(2*(a+1)/2).