TITLE: How to compute $\lim_{t \to \infty} \int_0^1 \sqrt{1+t^2 x^{2t-2}} dx$ without using arclength?
QUESTION [5 upvotes]: In computing
$$\lim_{t \to \infty} \int_0^1 \sqrt{1+t^2 x^{2t-2}} dx,$$
realizing the integral is of the form $\int_0^1 \sqrt{1+f'(x)^2}dx$ with $f(x)=x^t$ allows one to use the triangle inequality to make estimates on the arclength of $f(x)=x^t$ as $t \to \infty$. From this one can conclude the limit is 2.
How can I do this without thinking of this as an arclength? What are other methods of evaluation?

REPLY [2 votes]: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$Laplace\ Asymptotic\ Method:$

\begin{align}
&\color{#f00}{\lim_{t \to \infty}\int_{0}^{1}\root{1 + t^{2}x^{2t - 2}}\,\dd x} =
\lim_{t \to \infty}\int_{0}^{1}\root{1 + t^{2}\pars{1 - x}^{2t - 2}}\,\dd x
\\[5mm] = &\
\lim_{t \to \infty}\int_{0}^{1}
\exp\pars{{1 \over 2}\ln\pars{1 + t^{2}\bracks{1 - x}^{2t - 2}}}\,\dd x
\\[5mm] = &\
\lim_{t \to \infty}\int_{0}^{\infty}
\exp\pars{{1 \over 2}\ln\pars{1 + t^{2}} -
{t^{3} - t^{2} \over 2\pars{1 + t^{2}}}\, x}\,\dd x =
\lim_{t \to \infty}\bracks{%
\root{1 + t^{2}}\,{2\pars{1 + t^{2}} \over \pars{t - 1}t^{2}}} = \color{#f00}{2}
\end{align}