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Question Number 33594 by abdo imad last updated on 19/Apr/18

$$calculate{lim}_{x\to 1^{-}}\frac{1}{{(1-x)}^{\alpha }}(arcsinx-\frac{\pi }{2}).$$

Commented by abdo imad last updated on 24/Apr/18

$$letusethech.1-x=t\Rightarrow {lim}_{x\to 1^{-}}\frac{1}{{(1-x)}^{\alpha }}(arcsinx-\frac{\pi }{2})$$$$={lim}_{t\to 0^{+}}\frac{1}{t^{\alpha }}(arcsin(1-t)-\frac{\pi }{2})$$$$={lim}_{t\to 0^{+}}\frac{arcsin(1-t)-\frac{\pi }{2}}{t^{\alpha }}letusehospitaltheorem$$$$u\left(t\right)=arcsin(1-t)-\frac{\pi }{2}andv\left(t\right)=t^{\alpha }$$$$u^{{}^{\prime }}\left(t\right)=\frac{-1}{\sqrt{1-{(1-t)}^2}}and{v}^{{}^{\prime }}\left(t\right)=\alpha t^{\alpha -1}\Rightarrow$$$$u^{{}^{\prime }}\left(t\right)=\frac{-1}{\sqrt{1-t^2+2t-1}}=-\frac{1}{\sqrt{2t-t^2}}if\alpha >1lim\alpha t^{\alpha -1}=0^{+}$$$${lim}_{t\to 0^{+}}{u}^{{}^{\prime }}\left(t\right)=-\mathrm{\infty }\Rightarrow lim{}_{t\to o^{+}}\frac{arcsin(1-t)-\frac{\pi }{2}}{t^{\alpha }}=-\mathrm{\infty }$$$$if0<\alpha <1wedothech.\alpha =\frac{1}{\lambda }with\lambda >1\Rightarrow$$$${lim}_{t\to 0^{+}}\frac{arcsin(1-t)-\frac{\pi }{2}}{t^{\alpha }}={lim}_{t\to 0^{+}}\frac{arcsin(1-t)-\frac{\pi }{2}}{t^{\frac{1}{\lambda }}}$$$$={lim}_{t\to 0^{+}}\frac{-1}{\frac{1}{\lambda }{t}^{\frac{1}{\lambda }-1}\sqrt{2t-t^2}}={lim}_{t\to 0^{+}}\frac{-\lambda }{\sqrt{2t-t^2}}{t}^{1-\frac{1}{\lambda }}for$$$$thatwemustcalculate{u}^{{}^{″}}\left(t\right)and{v}^{{}^{″}}\left(t\right)...becontinued....$$

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