\displaystyle
\lim_{x\to0}\left((\cos x)+ax^2+
bx^3+
cx^4\right)^{x^{-2}}
.\exp(N/2)
",
where N
is an integer.(x)=e^x
.
exp(
2*a-1/2)
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})
.