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Question Number 111498 by bemath last updated on 04/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\left(1\right)\underset{x\to \mathrm{\infty }}{\lim }\sqrt[3]{{2\mathrm{x}}^2-{\mathrm{x}}^3}+\mathrm{x}?$$$$\left(2\right)\underset{x\to 1}{\lim }{\left(\frac{1}{\mathrm{x}}\right)}^{\frac{1}{\sin \pi \mathrm{x}}}?$$$$\left(3\right)\underset{0}{\stackrel{{\mathrm{x}}^2}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{x}\cos \left(\pi \mathrm{x}\right).\mathrm{Find}\mathrm{f}\left(4\right).$$

Answered by john santu last updated on 04/Sep/20

$$letx=\frac{1}{t}\to \{\begin{array}{l}x\to \mathrm{\infty }\\ t\to 0\end{array}$$$$L=\underset{t\to 0}{\lim }\sqrt[3]{\frac{2}{t^2}-\frac{1}{t^3}}+\frac{1}{t}$$$$L=\underset{t\to 0}{\lim }\sqrt[3]{\frac{2t-1}{t^3}}+\frac{1}{t}$$$$L=\underset{t\to 0}{\lim }\frac{\sqrt[3]{2t-1}+1}{t}$$$$L=\underset{t\to 0}{\lim }\frac{2t}{t[{(2t-1)}^2-(2t-1)+1]}$$$$L=\underset{t\to 0}{\lim }\frac{2}{{(2t-1)}^2+2-2t}=\frac{2}{3}.$$

Answered by bemath last updated on 04/Sep/20

Commented by bemath last updated on 04/Sep/20

$$typoin\sqrt[3]{x}itshouldbe\sqrt[3]{x^3}$$

Answered by john santu last updated on 04/Sep/20

$$\left(2\right)\ln L=\underset{x\to 1}{\lim }\left(\frac{-\ln x}{\sin \pi x}\right)$$$$\ln L=\underset{x\to 1}{\lim }\left(\frac{-\frac{1}{x}}{\pi \cos \pi x}\right)$$$$\ln L=\frac{1}{\pi }\Rightarrow L=e^{\frac{1}{\pi }}=\sqrt[\pi ]{e}$$

Answered by john santu last updated on 04/Sep/20

$$\left(3\right)\frac{d}{dx}\left[{\int }_0^{x^2}f\left(t\right)dt\right]=\frac{d}{dx}\left[x\cos \pi x\right]$$$$\Rightarrow 2xf\left(x^2\right)=\cos \pi x-\pi x\sin \pi x$$$$putx=2$$$$\Rightarrow 4f\left(4\right)=\cos 2\pi -2\pi \sin 2\pi$$$$\Rightarrow f\left(4\right)=\frac{1}{4}$$

Answered by mathmax by abdo last updated on 04/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right){=}^3\sqrt{-{\mathrm{x}}^3+{2\mathrm{x}}^2}+\mathrm{x}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}{-}^3\sqrt{{\mathrm{x}}^3(1-\frac{2}{\mathrm{x}})}$$$$=\mathrm{x}-\mathrm{x}{(1-\frac{2}{\mathrm{x}})}^{\frac{1}{3}}\sim \mathrm{x}-\mathrm{x}(1-\frac{2}{3\mathrm{x}})\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{2}{3}\Rightarrow {\lim }_{\mathrm{x}\to \mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{3}$$

Answered by mathmax by abdo last updated on 04/Sep/20

$$2)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\left(\frac{1}{\mathrm{x}}\right)}^{\frac{1}{\sin \left(\pi \mathrm{x}\right)}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{-\mathrm{lnx}}{\sin \left(\pi \mathrm{x}\right)}}\mathrm{we}\mathrm{have}$$$${\lim }_{\mathrm{x}\to 1}\frac{\mathrm{lnx}}{\sin \left(\pi \mathrm{x}\right)}{=}_{\mathrm{hospital}}{\lim }_{\mathrm{x}\to 1}\frac{1}{\mathrm{x}(\pi \cos \left(\pi \mathrm{x}\right)}=-\frac{1}{\pi }\Rightarrow$$$${\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{1}{\pi }}{=}^{\pi }\sqrt{\mathrm{e}}$$

Commented by bobhans last updated on 04/Sep/20

$$\mathrm{typo}{\mathrm{e}}^{-\frac{1}{\pi }}=\frac{1}{{\mathrm{e}}^{\frac{1}{\pi }}}$$

Commented by mathmax by abdo last updated on 04/Sep/20

$$\mathrm{no}\mathrm{typo}!$$

Commented by bobhans last updated on 04/Sep/20

$$\mathrm{why}?\mathrm{you}\mathrm{got}{\mathrm{e}}^{-\frac{1}{\pi }}?\mathrm{it}\mathrm{not}\mathrm{same}\mathrm{to}$$$$\sqrt[\pi ]{\mathrm{e}}???$$

Answered by mathmax by abdo last updated on 04/Sep/20

$$3)\mathrm{we}\mathrm{have}{\int }_0^{{\mathrm{x}}^2}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{xcos}\left(\pi \mathrm{x}\right)\mathrm{let}\mathrm{derivate}\Rightarrow$$$$2\mathrm{xf}\left({\mathrm{x}}^2\right)=\cos \left(\pi \mathrm{x}\right)-\pi \mathrm{xsin}\left(\pi \mathrm{x}\right)$$$$\mathrm{x}=2\mathrm{we}\mathrm{get}4\mathrm{f}\left(4\right)=\cos \left(2\pi \right)-2\pi \sin \left(2\pi \right)=1\Rightarrow \mathrm{f}\left(4\right)=\frac{1}{4}$$

Answered by Dwaipayan Shikari last updated on 04/Sep/20

$$\underset{x\to 1}{\lim }{\left(\frac{1}{x}\right)}^{\frac{1}{sin\pi x}}=y$$$$\frac{1}{sin\pi x}log\left(\frac{1}{x}\right)=logy$$$$\frac{\frac{1}{x}-1}{sin\pi x}\frac{log(1+\frac{1}{x}-1)}{(\frac{1}{x}-1)}=logy$$$$\underset{w\to 0}{\lim }\frac{\frac{1}{w+1}-1}{sin(\pi w+\pi )}=logy$$$$\underset{w\to 0}{\lim }\frac{\frac{-w}{w+1}}{-sin\pi w}=logy$$$$\underset{x\to 0}{\mathrm{li}m}\frac{\frac{-w}{w+1}}{-w\pi }=logy$$$$\frac{1}{\pi }=logy$$$$y=e^{\frac{1}{\pi }}$$

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