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Question Number 94679 by i jagooll last updated on 20/May/20

Commented by i jagooll last updated on 20/May/20

$$\mathrm{dear}\mathrm{tinkutara}\mathrm{sir}\mathrm{how}\mathrm{my}\mathrm{image}\mathrm{post}?$$

Commented by i jagooll last updated on 20/May/20

$$\mathrm{thank}\mathrm{a}\mathrm{lot}\mathrm{of}\mathrm{sir}$$

Answered by Ar Brandon last updated on 20/May/20

$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\frac{\pi }{2}\right)}{{\mathrm{x}}^2}-\underset{\mathrm{x}=0}{\lim }\frac{{\cos }^2\mathrm{x}}{{\mathrm{x}}^2}=0+\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinxcosx}}{\mathrm{x}}$$$$=1\times 1=1$$

Commented by i jagooll last updated on 20/May/20

cool man ����

Answered by mathmax by abdo last updated on 20/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \left(\frac{\pi }{2}\right)-{\cos }^2\mathrm{x}}{{\mathrm{x}}^2}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1-{\cos }^2\mathrm{x}}{{\mathrm{x}}^2}$$$$\mathrm{we}\mathrm{hsve}\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow {\cos }^2\mathrm{x}\sim 1-{\mathrm{x}}^2+\frac{{\mathrm{x}}^4}{4}\Rightarrow 1-{\cos }^2\mathrm{x}\sim {\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{4}$$$$\frac{1-{\cos }^2\mathrm{x}}{{\mathrm{x}}^2}\sim 1-\frac{{\mathrm{x}}^2}{4}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=1$$

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