Answer to Question #131236 in Calculus for Moel Tariburu

Question

Evaluate the limits:
lim┬(x→0)⁡〖1/x^3 〗 ∫_0^x▒〖t^2/(t^4+1) dt〗

Answer 1

If $0\leq t\leq x$, then $1+t^4\leq 1+x^4$ , hence

$1=\frac{1}{1+0^4}\geq\frac{1}{1+t^4}\geq\frac{1}{1+x^4},$ hence

$\frac{1}{3}=\frac{1}{x^3}\intop_0^x t^2dt\geq\\ \frac{1}{x^3}\intop_0^x\frac{t^2}{1+t^4}dt\geq\frac{1}{x^3}\intop_0^x\frac{t^2}{1+x^4}dt=\frac{1}{3(1+x^4)}$,

$\lim\limits_{x\rightarrow+0}\frac{1}{3}\geq\\ \lim\limits_{x\rightarrow+0}\frac{1}{x^3}\intop_0^x\frac{1}{1+t^4}dt\geq\\ \lim\limits_{x\rightarrow+0}\frac{1}{3(1+x^4)}=\frac{1}{3(1+0)}=\frac{1}{3},$

By The Squeeze Theorem

$\lim\limits_{x\rightarrow+0}\frac{1}{x^3}\intop_0^x\frac{t^2}{1+t^4}dt=\frac{1}{3}.$

Similarly, if $x\leq t\leq 0$, then $1+t^4\leq 1+x^4$ ,

hence

$1=\frac{1}{1+0^4}\geq\frac{1}{1+t^4}\geq\frac{1}{1+x^4},$

hence for $x<0$

$\frac{1}{3}=\frac{1}{x^3}\intop_0^x t^2dt=-\frac{1}{x^3}\intop_x^0 t^2dt\geq\\ -\frac{1}{x^3}\intop_x^0\frac{t^2}{1+t^4}dt=\frac{1}{x^3}\intop_0^x\frac{t^2}{1+t^4}dt\geq-\frac{1}{x^3}\intop_x^0\frac{t^2}{1+x^4}dt=\\ \frac{1}{x^3}\intop_0^x\frac{t^2}{1+x^4}dt=\frac{1}{3(1+x^4)},$

hence

$\lim\limits_{x\rightarrow-0}\frac{1}{3}\geq\\ \lim\limits_{x\rightarrow-0}\frac{1}{x^3}\intop_0^x\frac{1}{1+t^4}dt\geq\\ \lim\limits_{x\rightarrow-0}\frac{1}{3(1+x^4)}=\frac{1}{3(1+0)}=\frac{1}{3},$

by The Squeeze Theorem

$\lim\limits_{x\rightarrow-0}\frac{1}{x^3}\intop_0^x\frac{t^2}{1+t^4}dt=\frac{1}{3},$

hence

$\lim\limits_{x\rightarrow0}\frac{1}{x^3}\intop_0^x\frac{1}{1+t^4}dt=\frac{1}{3}.$

Answer: $\frac{1}{3}.$