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Question Number 96232 by joki last updated on 30/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\cos \sqrt{x}-\cos x}{1-\cos \sqrt{x}}=$$

Commented by bobhans last updated on 31/May/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\cos \sqrt{\mathrm{x}}-\cos \mathrm{x}}{1-\cos \sqrt{\mathrm{x}}}=\mathrm{does}\mathrm{not}\mathrm{exist}!$$

Answered by behi83417@gmail.com last updated on 30/May/20

$$\mathrm{x}=\frac{1}{\mathrm{t}}\Rightarrow [\mathrm{x}\to \mathrm{\infty }\Rightarrow \mathrm{t}\to 0]$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{\cos \sqrt{\frac{1}{\mathrm{t}}}-\mathrm{cost}}{1-\cos \sqrt{\frac{1}{\mathrm{t}}}}=\underset{\mathrm{t}\to 0}{\lim }\frac{(1-\frac{1}{2\mathrm{t}})-(1-\frac{{\mathrm{t}}^2}{2})}{\frac{1}{2\mathrm{t}}}=$$$$=\underset{\mathrm{t}\to 0}{\lim }\frac{\frac{{\mathrm{t}}^2}{2}-\frac{1}{2\mathrm{t}}}{\frac{1}{2\mathrm{t}}}=\underset{\mathrm{t}\to 0}{\lim }\frac{{\mathrm{t}}^3-1}{1}=-1.◼$$

Commented by bobhans last updated on 31/May/20

$$\mathrm{no}.\mathrm{it}\mathrm{wrong}$$

Answered by mathmax by abdo last updated on 31/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(\sqrt{\mathrm{x}}\right)-\mathrm{cosx}}{1-\cos \left(\sqrt{\mathrm{x}}\right)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\cos \left(\sqrt{\mathrm{x}}\right)-1+1-\mathrm{cosx}}{1-\cos \left(\sqrt{\mathrm{x}}\right)}$$$$=-1+\frac{1-\mathrm{cosx}}{1-\cos \left(\sqrt{\mathrm{x}}\right)}=-1+\mathrm{g}\left(\mathrm{x}\right)\mathrm{with}\mathrm{g}\left(\mathrm{x}\right)=\frac{1-\cos \left(\mathrm{x}\right)}{1-\cos \left(\sqrt{\mathrm{x}}\right)}$$$$\mathrm{let}{\mathrm{x}}_{\mathrm{n}}={\mathrm{n}}^2{\pi }^2\mathrm{we}\mathrm{have}{\mathrm{x}}_{\mathrm{n}}\to \mathrm{\infty }\mathrm{and}\mathrm{g}\left({\mathrm{x}}_{\mathrm{n}}\right)=\frac{1-\cos \left({\mathrm{n}}^2{\pi }^2\right)}{1-\cos \left(\mathrm{n}\pi \right)}$$$$=\frac{1-\cos \left({\mathrm{n}}^2{\pi }^2\right)}{1-{(-1)}^{\mathrm{n}}}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\mathrm{g}\left({\mathrm{x}}_{\mathrm{n}}\right)\mathrm{dont}\mathrm{exist}...$$

Commented by joki last updated on 31/May/20

$$answeris1?$$

Commented by mathmax by abdo last updated on 31/May/20

$$\mathrm{show}\mathrm{your}\mathrm{work}\mathrm{the}\mathrm{answer}\mathrm{given}\mathrm{by}\mathrm{sir}\mathrm{behi}\mathrm{is}\mathrm{not}\mathrm{correct}...$$

Commented by joki last updated on 31/May/20

$$notsolution\left(\mathrm{\infty }\right)?$$

Answered by bemath last updated on 31/May/20

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