[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
Evaluation of a limit of the form
[math]\lim_{x\goesto a} f(x)^{g(x)}[/math]
is not obvious if any one of the following
three possibilities occurs.
(i) \quad [math]\lim_{x\goesto a} f(x) = \lim_{x\goesto a} g(x) = 0[/math].
(ii) \quad [math]\lim_{x\goesto a} f(x) = 1[/math] and
[math]\lim_{x\goesto a} g(x) = \infty[/math].
(iii) \quad [math]\lim_{x\goesto a} f(x) = \infty[/math] and
[math]\lim_{x\goesto a} g(x) = 0[/math].
These three types are usually referred to,
respectively, as the indeterminate forms
[math]0^0[/math], [math]1^\infty[/math], and [math]\infty^0[/math].
The standard attack, akin to logarithmic differentiation,
is the following: Let
[[math]]
h(x) = \ln f(x)^{g(x)} = g(x) \ln f(x) =
\frac{\ln f(x)}{\frac1{g(x)}}
.
[[/math]]
One then applies L'H\^opital's Rule to the quotient,
thereby hopefully discovering that
[math]\lim_{x\goesto a} h(x)[/math] exists and what its value is.
If it does exist, it follows by the continuity
of the exponential function that
[[math]]
e^{\left[ \lim_{x\goesto a} h(x)\right]} =
\lim_{x\goesto a} e^{h(x)}
.
[[/math]]
But, since
[[math]]
e^{h(x)} = e^{\ln f(x)^{g(x)}} = f(x)^{g(x)}
,
[[/math]]
we therefore conclude that
[[math]]
\lim_{x\goesto a} f(x)^{g(x)} =
e^{\left[ \lim_{x\goesto a} h(x)\right]}
,
[[/math]]
and the problem is solved.
Apply this method to evaluate the following limits.
- [math]\lim_{x\goesto0+} x^x[/math]
- [math]\lim_{x\goesto\infty} x^{\frac1x}[/math]
- [math]\lim_{x\goesto0} (1+2x)^{\frac1x}[/math].