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

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

$$\underset{x\to 0}{\lim }{\int }_0^x\frac{t^{\mu -1}}{x^{\mu }}u\left(t\right)dt$$$$=\underset{x\to 0}{\lim }\frac{{\int }_0^x{t}^{\mu -1}u\left(t\right)dt}{x^{\mu }}\left(x^{\mu }isindependentfromtinintegral\right)$$$$=\underset{x\to 0}{\lim }\frac{{\left({\int }_0^x{t}^{\mu -1}u\left(t\right)dt\right)}_{w.r.t.x}^{{}^{\prime }}}{{\left(x^{\mu }\right)}_{w.r.t.x}^{{}^{\prime }}}\left(applying{l}^{\prime }hopitalrule\right)$$$$=\underset{x\to 0}{\lim }\frac{x^{\mu -1}u\left(x\right)}{\mu x^{\mu -1}}$$$$=\underset{x\to 0}{\lim }\frac{u\left(x\right)}{\mu }$$$$=\frac{u\left(0\right)}{\mu }$$

Commented by ghakhan88 last updated on 17/Nov/21

$$nice.thnx$$

Commented by Tawa11 last updated on 17/Nov/21

$$\mathrm{Great}\mathrm{sir}.$$$$\mathrm{Please}\mathrm{sir},\mathrm{help}\mathrm{me}\mathrm{check}\mathrm{the}\mathrm{sequence}\mathrm{on}\mathrm{Q}159488$$

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