Let L=\displaystyle \lim_{x\to0}\left((\cos x)+ax^2+ bx^3+ cx^4\right)^{x^{-2}} .
Then L=\displaystyle \lim_{x\to0}\exp \left(\ln\left[\left((\cos x)+ax^2+ bx^3+ cx^4\right)^{x^{-2}}\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=(\cos x)+ax^2+ bx^3+ cx^4    and     B=x^{-2}   .

We get L=\displaystyle \lim_{x\to0}\exp \left(\left[x^{-2}\right]\left[\ln\left((\cos x)+ax^2+ bx^3+ cx^4\right)\right]\right) .

Next, use that \lim and \exp commute.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-2}\right]\left[\ln\left((\cos x)+ax^2+ bx^3+ cx^4\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+(\cos x)+ax^2+ bx^3+ cx^4.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-2}\right]\left[-1+(\cos x)+ax^2+ bx^3+ cx^4\right]\right) .

Because
    \cos x=1-\frac{x^2}{2}+\frac{x^4}{6}+\cdots
and because
    a power series is asymptotic (as x\to0) to its lowest order term,
it follows that
    -1+(\cos x)+ax^2+ bx^3+ cx^4\quad\sim\quad\frac{2*a-1}{2}x^2,\quad \hbox{as }x\to0.

Then L=\displaystyle \exp\left(\lim_{x\to0} \left[x^{-2}\right]\left[\frac{2*a-1}{2}x^2\right]\right) .

Then L=\displaystyle \exp\left(\lim_{x\to0}\,\,\frac{2*a-1}{2}\right) ,   because
   \left[x^{-2}\right]\left[\frac{2*a-1}{2}x^2\right]= \frac{2*a-1}{2}\hbox{ on }x\ne0.

Then L=\displaystyle \exp\left(\frac{2*a-1}{2}\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(\frac{2*a-1}{2}).