Question-159431

Question Number 159431 by ghakhan88 last updated on 16/Nov/21

Answered by mr W last updated on 17/Nov/21

lim_(x→0) ∫_0 ^x (t^(μ−1) /x^μ )u(t)dt =lim_(x→0) ((∫_0 ^x t^(μ−1) u(t)dt)/x^μ ) (x^μ is independent from t in integral) =lim_(x→0) (((∫_0 ^x t^(μ−1) u(t)dt)_(w.r.t. x) ^′ )/((x^μ )_(w.r.t. x) ^′ )) (applying l′hopital rule) =lim_(x→0) ((x^(μ−1) u(x))/(μx^(μ−1) )) =lim_(x→0) ((u(x))/μ) =((u(0))/μ)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\int_{\mathrm{0}} ^{{x}} \frac{{t}^{\mu−\mathrm{1}} }{{x}^{\mu} }{u}\left({t}\right){dt} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{0}} ^{{x}} {t}^{\mu−\mathrm{1}} {u}\left({t}\right){dt}}{{x}^{\mu} }\:\:\:\:\:\left({x}^{\mu} \:{is}\:{independent}\:{from}\:{t}\:{in}\:{integral}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\int_{\mathrm{0}} ^{{x}} {t}^{\mu−\mathrm{1}} {u}\left({t}\right){dt}\right)_{{w}.{r}.{t}.\:{x}} ^{'} }{\left({x}^{\mu} \right)_{{w}.{r}.{t}.\:{x}} ^{'} }\:\:\:\left({applying}\:{l}'{hopital}\:{rule}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mu−\mathrm{1}} {u}\left({x}\right)}{\mu{x}^{\mu−\mathrm{1}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{u}\left({x}\right)}{\mu}\:\:\: \\ $$$$=\frac{{u}\left(\mathrm{0}\right)}{\mu}\: \\ $$

Commented by ghakhan88 last updated on 17/Nov/21

$${nice}.\:{thnx} \\ $$

Commented by Tawa11 last updated on 17/Nov/21

Great sir. Please sir, help me check the sequence on Q159488

$$\mathrm{Great}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{on}\:\:\:\:\mathrm{Q159488} \\ $$

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