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Question Number 51005 by Tawa1 last updated on 23/Dec/18

Commented by maxmathsup by imad last updated on 23/Dec/18

$$letA\left(\theta \right)=\frac{sin(\theta -\frac{\pi }{6})}{\sqrt{3}-2cos\theta }changement\theta -\frac{\pi }{6}=xgive$$$$A\left(\theta \right)=\frac{sinx}{\sqrt{3}-2cos(x+\frac{\pi }{6})}=\frac{sinx}{\sqrt{3}-2(cosxcos\left(\frac{\pi }{6}\right)-sinxsin\left(\frac{\pi }{6}\right))}$$$$=\frac{sinx}{\sqrt{3}-2(\frac{\sqrt{3}}{2}cos-\frac{1}{2}sinx)}=\frac{sinx}{\sqrt{3}-\sqrt{3}cosx+sinx}but{lim}_{\theta \to \frac{\pi }{6}}={lim}_{x\to 0}\frac{sinx}{\sqrt{3}-\sqrt{3}cosx+sinx}$$$$sinx\sim xandcosx\sim 1-\frac{x^2}{2}\Rightarrow \frac{sinx}{\sqrt{3}-\sqrt{3}cosx-sinx}\sim \frac{x}{\sqrt{3}-\sqrt{3}(1-\frac{x^2}{2})-x}$$$$=\frac{x}{\frac{\sqrt{3}}{2}{x}^2-x}=\frac{1}{\frac{\sqrt{3}}{2}x-1}\to -1(x\to 0)\Rightarrow {lim}_{\theta \to \frac{\pi }{6}}A\left(\theta \right)=-1.$$$$erroroftypo\frac{sinx}{\sqrt{3}-\sqrt{3}cosx+sinx}\sim \frac{x}{\sqrt{3}-\sqrt{3}(1-\frac{x^2}{2})+x}=\frac{x}{\frac{\sqrt{3}}{2}{x}^2+x}=$$$$=\frac{1}{\frac{\sqrt{3}}{2}x+1}\to 1(x\to 0)\Rightarrow ★{lim}_{\theta \to \frac{\pi }{6}}A\left(\theta \right)=1★$$

Commented by maxmathsup by imad last updated on 23/Dec/18

$$letA\left(x\right)=\frac{1-cosx\sqrt{cos\left(2x\right)}}{x^2}wehavex\in V\left(0\right)$$$$cosx\sim 1-\frac{x^2}{2}andcos\left(2x\right)\sim 1-\frac{{\left(2x\right)}^2}{2}=1-2x^2\Rightarrow \sqrt{cos\left(2x\right)}\sim \sqrt{1-2x^2}$$$$\sim 1+\frac{1}{2}(-2x^2)=1-x^2\Rightarrow A\left(x\right)\sim \frac{1-(1-\frac{x^2}{2})(1-x^2)}{x^2}$$$$=\frac{1-(1-x^2-\frac{x^2}{2}+\frac{x^4}{2})}{x^2}=\frac{x^2+\frac{x^2}{2}-\frac{x^4}{2}}{x^2}=\frac{3}{2}-\frac{x^2}{2}\to \frac{3}{2}(x\to 0)\Rightarrow$$$${lim}_{x\to 0}A\left(x\right)=\frac{3}{2}.$$

Commented by Tawa1 last updated on 24/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$2)t=\theta -\frac{\pi }{6}$$$$li\underset{t\to 0}{m}\frac{sin\left(t\right)}{\sqrt{3}-2cos(t+\frac{\pi }{6})}$$$$=li\underset{t\to 0}{m}\frac{sint}{\sqrt{3}-2(cost.\frac{\sqrt{3}}{2}-sint.\frac{1}{2})}$$$$=li\underset{t\to 0}{m}\frac{sint}{\sqrt{3}-\sqrt{3}cost+sint}$$$$=li\underset{t\to 0}{m}\frac{2sin\frac{t}{2}cos\frac{t}{2}}{\sqrt{3}\times 2{sin}^2\frac{t}{2}+2sin\frac{t}{2}cos\frac{t}{2}}$$$$=li\underset{t\to 0}{m}\frac{cos\frac{t}{2}}{\sqrt{3}sin\frac{t}{2}+cos\frac{t}{2}}=1$$$$ormethod$$$$li\underset{\theta \to \frac{\pi }{6}}{m}\frac{sin(\theta -\frac{\pi }{6})}{2[\frac{\sqrt{3}}{2}-cos\left(\theta \right)]}[\frac{\sqrt{3}}{2}=cos\frac{\pi }{6}]$$$$\frac{1}{2}li\underset{\theta \to \frac{\pi }{6}}{m}\frac{2sin\left(\frac{\theta -\frac{\pi }{6}}{2}\right)cos\left(\frac{\theta -\frac{\pi }{6}}{2}\right)}{2sin\left(\frac{\frac{\pi }{6}+\theta }{2}\right)sin\left(\frac{\theta -\frac{\pi }{6}}{2}\right)}$$$$=\frac{1}{2}\times \frac{1}{\frac{1}{2}}=1$$

Commented by Tawa1 last updated on 23/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$mostwelcome...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$3)\underset{x\to 0}{\lim }\frac{1-cosx\sqrt{cos2x}}{x^2}$$$$\underset{x\to 0}{\lim }\frac{1-{cos}^{2}x\left(\frac{1-{tan}^{2}x}{1+{tan}^{2}x}\right)}{x^2(1+cosx\sqrt{cos2x})}$$$$=\underset{x\to 0}{\lim }\frac{{sec}^{2}x-{cos}^{2}x+{sin}^{2}x}{{sec}^{2}x\times x^2\times (1+cosx\sqrt{cos2x})}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }\frac{\frac{1}{{cos}^{2}x}-{cos}^{2}x+{sin}^{2}x}{x^2}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{{sin}^{2}x.(1+{cos}^{2}x)+{sin}^{2}x}{x^2{cos}^{2}x}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }{\left(\frac{sinx}{x}\right)}^2\times \left(\frac{2+{cos}^{2}x}{{cos}^{2}x}\right)$$$$=\frac{1}{2}\times 1\times \frac{3}{2}=\frac{3}{4}$$

Commented by Tawa1 last updated on 23/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$1)\underset{y\to 0}{\lim }\frac{x\{sec(x+y)-secx\}+ysec(x+y)}{y}$$$$=x\left[\underset{y\to 0}{\lim }\frac{sec(x+y)-secx}{y}\right]+secx$$$$=x\left[\underset{y\to 0}{\lim }\frac{cosx-cos(x+y)}{ycosxcos(x+y)}\right]+secx$$$$=x\left[\underset{y\to 0}{\lim }\frac{2sin(x+\frac{y}{2})sin\left(\frac{y}{2}\right)}{cosxcos(x+y)\times \frac{y}{2}\times 2}\right]+secx$$$$=x\times \frac{sinx}{{cos}^{2}x}+secx$$$$=xtanxsecx+secx$$

Commented by Tawa1 last updated on 23/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$$mostwelcome...$$

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