Question Number 159431 by ghakhan88 last updated on 16/Nov/21
Answered by mr W last updated on 17/Nov/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{t}^{\mu−\mathrm{1}} }{{x}^{\mu} }{u}\left({t}\right){dt} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{{x}} {t}^{\mu−\mathrm{1}} {u}\left({t}\right){dt}}{{x}^{\mu} }\:\:\:\:\:\left({x}^{\mu} \:{is}\:{independent}\:{from}\:{t}\:{in}\:{integral}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\int_{\mathrm{0}} ^{{x}} {t}^{\mu−\mathrm{1}} {u}\left({t}\right){dt}\right)_{{w}.{r}.{t}.\:{x}} ^{'} }{\left({x}^{\mu} \right)_{{w}.{r}.{t}.\:{x}} ^{'} }\:\:\:\left({applying}\:{l}'{hopital}\:{rule}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mu−\mathrm{1}} {u}\left({x}\right)}{\mu{x}^{\mu−\mathrm{1}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}\left({x}\right)}{\mu}\:\:\: \\ $$$$=\frac{{u}\left(\mathrm{0}\right)}{\mu}\: \\ $$
Commented by ghakhan88 last updated on 17/Nov/21
$${nice}.\:{thnx} \\ $$
Commented by Tawa11 last updated on 17/Nov/21
$$\mathrm{Great}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{on}\:\:\:\:\mathrm{Q159488} \\ $$