\displaystyle
\lim_{x\to0}\left(1+ax^{k}+
bx^{k+1}+
cx^{k+2}\right)^{x^{-k}}
.\exp(N/2)
",
where N
is an integer.(x)=e^x
.
exp(
2*a/2)
Let L=\displaystyle
\lim_{x\to0}\left(1+ax^{k}+
bx^{k+1}+
cx^{k+2}\right)^{x^{-k}}
.
Then L=\displaystyle
\lim_{x\to0}\exp
\left(\ln\left[\left(1+ax^{k}+
bx^{k+1}+
cx^{k+2}\right)^{x^{-k}}\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}\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}\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}\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}\right]\left[ax^{k}\right]\right)
.
Then L=\displaystyle
\exp\left(\lim_{x\to0}\,\,a\right)
,
because \left[x^{-k}\right]\left[ax^{k}\right]=a
\hbox{ on }x\ne0
.
Then L=\displaystyle
\exp\left(a\right)
, because a limit of a constant is the constant.
Now enter 2*a
into the Answer box,
to indicate that the answer is \exp(2*a/2)
.