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Question Number 118427 by bramlexs22 last updated on 17/Oct/20

$$\left(1\right)\underset{x\to 0}{\lim }\sqrt[x^2]{\cos x}$$$$\left(2\right)\underset{x\to 1}{\lim }\frac{1+\cos \pi x}{x^2-2x+1}$$

Answered by Dwaipayan Shikari last updated on 17/Oct/20

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

Commented by Dwaipayan Shikari last updated on 17/Oct/20

$$Anotherway$$$${\left(cosx\right)}^{\frac{1}{x^2}}=y$$$$\underset{x\to 0}{\lim }\frac{1}{x^2}log\left(cosx\right)=logy$$$$\underset{x\to 0}{\lim }\frac{1}{x^2}log(1+cosx-1)=logy$$$$\underset{x\to 0}{\lim }\frac{cosx-1}{x^2}=logy\underset{x\to 0}{\lim }log(1+x)=x$$$$\underset{x\to 0}{\lim }\frac{-2{sin}^2\frac{x}{2}}{x^2}=logy$$$$logy=-\frac{1}{2}\Rightarrow y=\frac{1}{\sqrt{e}}$$

Answered by Dwaipayan Shikari last updated on 17/Oct/20

$$\underset{x\to 1}{\lim }\frac{2{cos}^2\frac{\pi x}{2}}{{(x-1)}^2}=\underset{x\to 1}{\lim }\frac{2{sin}^2(\frac{\pi }{2}-\frac{\pi }{2}x)}{{(x-1)}^2}$$$$=\underset{x\to 1}{\lim 2}.\frac{\left(\frac{{\pi }^2}{4}\right){(1-x)}^2}{{(x-1)}^2}=\frac{{\pi }^2}{2}$$

Answered by TANMAY PANACEA last updated on 17/Oct/20

$$2)\underset{x\to 1}{\lim }\frac{2{cos}^2\frac{\pi x}{2}}{{(x-1)}^2}$$$$cos\left(\frac{\pi x}{2}\right)=sin(\frac{\pi }{2}-\frac{\pi x}{2})=sin\frac{\pi }{2}(1-x)$$$$\underset{x\to 1}{\lim }\frac{2{sin}^2\frac{\pi }{2}(1-x)}{{(1-x)}^2\times \frac{{\pi }^2}{4}}\times \frac{{\pi }^2}{4}$$$$\theta =\frac{\pi }{2}(1-x)\to asx\to 1\theta \to 0$$$$\underset{\theta \to 0}{\lim }\frac{2\times {sin}^2\theta }{{\theta }^2}\times \frac{{\pi }^2}{4}=\frac{{\pi }^2}{2}$$

Answered by benjo_mathlover last updated on 17/Oct/20

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

Answered by benjo_mathlover last updated on 17/Oct/20

$$\left(2\right)setx=s+1\wedge s\to 0$$$$\underset{s\to 0}{\lim }\frac{1+\cos (\pi s+s)}{s^2}=\underset{s\to 0}{\lim }\frac{1-\cos \left(\pi s\right)}{s^2}$$$$\underset{s\to 0}{\lim }\frac{1-(1-\frac{{\pi }^2{s}^2}{2})}{s^2}=\frac{{\pi }^2}{2}$$

Answered by Bird last updated on 17/Oct/20

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

Answered by mathmax by abdo last updated on 17/Oct/20

$$2)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{1+\cos \left(\pi \mathrm{x}\right)}{{\mathrm{x}}^2-2\mathrm{x}+1}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{1+\cos \left(\pi \mathrm{x}\right)}{{(\mathrm{x}-1)}^2}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{chang}.$$$$\mathrm{x}-1=\mathrm{t}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(\mathrm{t}+1)=\frac{1+\cos \left(\pi (\mathrm{t}+1)\right)}{{\mathrm{t}}^2}$$$$=\frac{1-\cos \left(\pi \mathrm{t}\right)}{{\mathrm{t}}^2}/(\mathrm{x}\to 1\Rightarrow \mathrm{t}\to 0)\mathrm{so}1-\cos \left(\pi \mathrm{t}\right)\sim \frac{{\pi }^2{\mathrm{t}}^2}{2}\Rightarrow$$$$\frac{1-\cos \left(\pi \mathrm{t}\right)}{{\mathrm{t}}^2}\sim \frac{{\pi }^2}{2}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{f}(\mathrm{t}+1)=\frac{{\pi }^2}{2}={\lim }_{\mathrm{x}\to 1}\mathrm{f}\left(\mathrm{x}\right)$$

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