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Question Number 115780 by bemath last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{x^2}}=?$$$$\underset{x\to \mathrm{\infty }}{\lim }{(e^{3x}-5x)}^{\frac{1}{x}}=?$$$$\underset{x\to 0}{\lim }\frac{e^{2x}-2e^x+1}{\cos 3x-2\cos 2x+\cos x}=?$$$$\underset{x\to 0}{\lim }\frac{1}{x^2}-\cot {}^{2}x=?$$

Commented by bobhans last updated on 28/Sep/20

$$L=\underset{x\to 0}{\lim }{\left(\cos x\right)}^{\frac{1}{x^2}}=e^{\underset{x\to 0}{\lim }(\cos x-1).\frac{1}{x^2}}$$$$L=e^{\underset{x\to 0}{\lim }\frac{1-\frac{1}{2}{x}^2-1}{x^2}}=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }{\left(\mathrm{cosx}\right)}^{\frac{1}{{\mathrm{x}}^2}}={(1-\frac{{\mathrm{x}}^2}{2})}^{-\frac{2}{{\mathrm{x}}^2}.-\frac{1}{2}}={\mathrm{e}}^{-\frac{1}{2}}=\frac{1}{\sqrt{\mathrm{e}}}$$

Answered by bobhans last updated on 28/Sep/20

$$L=\underset{x\to \mathrm{\infty }}{\lim }{(e^{3x}-5x)}^{\frac{1}{x}}$$$$\ln L=\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (e^{3x}-5x)}{x}$$$$\ln L=\underset{x\to \mathrm{\infty }}{\lim }\frac{3e^{3x}-5}{e^{3x}-5x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{9e^{3x}}{3e^{3x}-5}$$$$\ln L=\underset{x\to \mathrm{\infty }}{\lim }\frac{9e^{3x}}{e^{3x}(3-\frac{5}{e^{3x}})}=3$$$$L=e^3$$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }\frac{1}{{\mathrm{x}}^2}-\frac{{\cos }^2\mathrm{x}}{{\sin }^2\mathrm{x}}=\frac{{\sin }^2\mathrm{x}-{\mathrm{x}}^2{\cos }^2\mathrm{x}}{{\mathrm{x}}^2{\sin }^2\mathrm{x}}=\frac{{\mathrm{x}}^2{(1-\frac{{\mathrm{x}}^2}{6})}^2-{\mathrm{x}}^2{(1-\frac{{\mathrm{x}}^2}{2})}^2}{{\mathrm{x}}^2{\sin }^2\mathrm{x}}$$$$=\frac{(2-\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^2}{6})(\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^2}{6})}{{\mathrm{x}}^2}=\frac{{\mathrm{x}}^2(2-\frac{{2\mathrm{x}}^2}{3})\left(\frac{1}{3}\right)}{{\mathrm{x}}^2}=\frac{2}{3}$$

Answered by Dwaipayan Shikari last updated on 28/Sep/20

$$\underset{x\to 0}{\lim }\frac{{({\mathrm{e}}^{\mathrm{x}}-1)}^2}{{\mathrm{x}}^2}.\frac{{\mathrm{x}}^2}{(\cos 3\mathrm{x}+\mathrm{cosx}-2\cos 2\mathrm{x})}$$$$=1.1.\frac{{\mathrm{x}}^2}{2(\cos 2\mathrm{xcosx}-\cos 2\mathrm{x})}=\frac{{\mathrm{x}}^2}{2\cos 2\mathrm{x}(\mathrm{cosx}-1)}$$$$=-\frac{1}{2}.\frac{{\mathrm{x}}^2}{{2\sin }^2\frac{\mathrm{x}}{2}}=-\frac{1}{4}.\frac{{\mathrm{x}}^2}{\frac{{\mathrm{x}}^2}{4}}=-1$$

Answered by mathmax by abdo last updated on 28/Sep/20

$${\left(\mathrm{cosx}\right)}^{\frac{1}{{\mathrm{x}}^2}}={\mathrm{e}}^{\frac{1}{{\mathrm{x}}^2}\ln \left(\mathrm{cosx}\right)}\mathrm{we}\mathrm{have}\ln \left(\mathrm{cosx}\right)\sim \ln (1-\frac{{\mathrm{x}}^2}{2})\sim -\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$\frac{1}{{\mathrm{x}}^2}\ln \left(\mathrm{cosx}\right)\sim -\frac{1}{2}\Rightarrow {\lim }_{\mathrm{x}\to 0}{\left(\mathrm{cosx}\right)}^{\frac{1}{{\mathrm{x}}^2}}={\mathrm{e}}^{-\frac{1}{2}}=\frac{1}{\sqrt{\mathrm{e}}}$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\mathrm{we}\mathrm{have}{({\mathrm{e}}^{3\mathrm{x}}-5\mathrm{x})}^{\frac{1}{\mathrm{x}}}=({\mathrm{e}}^{3\mathrm{x}}{(1-5\mathrm{x}{\mathrm{e}}^{-3\mathrm{x}})}^{\frac{1}{\mathrm{x}}}={\mathrm{e}}^3{(1-{5\mathrm{xe}}^{-3\mathrm{x}})}^{\frac{1}{\mathrm{x}}}$$$$={\mathrm{e}}^3{\mathrm{e}}^{\frac{1}{\mathrm{x}}\ln (1-{5\mathrm{xe}}^{-3\mathrm{x}})}\sim {\mathrm{e}}^3{\mathrm{e}}^{\frac{1}{\mathrm{x}}(-{5\mathrm{xe}}^{-3\mathrm{x}})}={\mathrm{e}}^3{\mathrm{e}}^{-{5\mathrm{e}}^{-3\mathrm{x}}}\Rightarrow {\mathrm{e}}^{-2}=\frac{1}{{\mathrm{e}}^2}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}{({\mathrm{e}}^{3\mathrm{x}}-5\mathrm{x})}^{\frac{1}{\mathrm{x}}}=\frac{1}{{\mathrm{e}}^2}$$

Commented by Dwaipayan Shikari last updated on 28/Sep/20

$${\mathrm{e}}^3{\mathrm{e}}^{-{5\mathrm{e}}^{-3\mathrm{x}}}={\mathrm{e}}^3.{\mathrm{e}}^{-5\left(0\right)}={\mathrm{e}}^3(\mathrm{As}{\mathrm{e}}^{-3\mathrm{x}}\to 0)$$

Commented by Bird last updated on 28/Sep/20

$$sooryitakelimitat0$$$${lim}_{x\to +\mathrm{\infty }}{(e^{3x}-5x)}^{\frac{1}{x}}=e^3.e^0=e^3$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}+1}{\cos \left(3\mathrm{x}\right)-2\cos \left(2\mathrm{x}\right)+\mathrm{cosx}}(\mathrm{x}\to 0)\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{1+2\mathrm{x}+{2\mathrm{x}}^2-2(1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2})+1}{1-\frac{{9\mathrm{x}}^2}{2}-2(1-{2\mathrm{x}}^2)+1-\frac{{\mathrm{x}}^2}{2}}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^2}{-\frac{{9\mathrm{x}}^2}{2}+{4\mathrm{x}}^2-\frac{{\mathrm{x}}^2}{2}}=\frac{{\mathrm{x}}^2}{-{5\mathrm{x}}^2+{4\mathrm{x}}^2}=\frac{{\mathrm{x}}^2}{-{\mathrm{x}}^2}=-1\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=-1$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{we}\mathrm{have}$$$${\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}+1}{\cos \left(3\mathrm{x}\right)-2\cos \left(2\mathrm{x}\right)+\mathrm{cosx}}={\lim }_{\mathrm{x}\to 0}\frac{{2\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}}{-3\sin \left(3\mathrm{x}\right)+4\sin \left(2\mathrm{x}\right)-\mathrm{sinx}}$$$$={\lim }_{\mathrm{x}\to 0}\frac{{4\mathrm{e}}^{2\mathrm{x}}-{2\mathrm{e}}^{\mathrm{x}}}{-9\cos \left(3\mathrm{x}\right)+8\cos \left(2\mathrm{x}\right)-\mathrm{cosx}}=\frac{2}{-9+8-1}=\frac{2}{-2}=-1$$

Answered by mathmax by abdo last updated on 28/Sep/20

$$\mathrm{we}\mathrm{have}\frac{1}{{\mathrm{x}}^2}-{\mathrm{cotan}}^2\mathrm{x}=\frac{1}{{\mathrm{x}}^2}-\frac{{\cos }^2\mathrm{x}}{{\sin }^2\mathrm{x}}=\frac{{\sin }^2\mathrm{x}-{\mathrm{x}}^2{\cos }^2\mathrm{x}}{{\mathrm{x}}^2{\sin }^2\mathrm{x}}$$$$=\frac{\frac{1-\cos \left(2\mathrm{x}\right)}{2}-{\mathrm{x}}^2\times \frac{1+\cos \left(2\mathrm{x}\right)}{2}}{{\mathrm{x}}^2.\frac{1-\cos \left(2\mathrm{x}\right)}{2}}=\frac{1}{2}.\frac{1-\cos \left(2\mathrm{x}\right)-{\mathrm{x}}^2(1+\cos \left(2\mathrm{x}\right))}{{\mathrm{x}}^2(1-\cos \left(2\mathrm{x}\right))}$$$$\Rightarrow \frac{1}{{\mathrm{x}}^2}-{\mathrm{cotan}}^2\mathrm{x}\sim \frac{1}{2}\frac{{2\mathrm{x}}^2-{\mathrm{x}}^2(1+1-{2\mathrm{x}}^2)}{{\mathrm{x}}^2\times \left({2\mathrm{x}}^2\right)}=\frac{1}{2}\times \frac{{2\mathrm{x}}^4}{{2\mathrm{x}}^4}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{1}{{\mathrm{x}}^2}-{\mathrm{cotan}}^2\mathrm{x}=\frac{1}{2}$$

Commented by Dwaipayan Shikari last updated on 28/Sep/20

Answered by john santu last updated on 29/Sep/20

$$\underset{x\to 0}{\lim }\frac{1}{x^2}-\frac{1}{\tan {}^{2}x}=\underset{x\to 0}{\lim }\frac{\tan {}^{2}x-x^2}{x^2\tan {}^{2}x}$$$$=\underset{x\to 0}{\lim }\frac{\tan x+x}{\tan x}\times \underset{x\to 0}{\lim }\frac{\tan x-x}{x^2\tan x}$$$$=\underset{x\to 0}{\lim }(1+\frac{x}{\tan x})\times \underset{x\to 0}{\lim }\frac{x+\frac{1}{3}{x}^3-x}{x^3}$$$$=2\times \underset{x\to 0}{\lim }\frac{\frac{1}{3}{x}^3}{x^3}=\frac{2}{3}$$

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