Question Number 176375 by cortano1 last updated on 17/Sep/22
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{3}}{\mathrm{x}+\sqrt{\mathrm{x}}}\:=?\: \\ $$
Commented by a.lgnaoui last updated on 18/Sep/22
 ((5(√X))/( (√X) +1))−(√X)sin X(((sin X +1)/( (√X) +1))) siut t=(√X) ((5t)/(t+1))−tsin (t^2 )(((sin (t^2 )+1)/(t+1))) ((5t)/(t+1))−(t/(t+1))[sin (t^2 )(sin (t^2 )+1] x→0+ t→+[∞ lim_(x→0+) =lim_(t→+∞) ((5t)/(t+1))−(t/(t+1))[sin (t^2 )(sin (t^2 )+1] =5−lim_(X→+∞) [sin(X)(sin (X)+1)] X=(2k+1)(π/2) sin X →1 donc lim_(x→0+) ((2cos^2 ((1/x))−sin ((1/x)))/(x+(√x)))=5−2=3](https://www.tinkutara.com/question/Q176431.png)
$${posons}\:\:{X}=\frac{\mathrm{1}}{{x}}\:\:\:{x}=\frac{\mathrm{1}}{{X}} \\ $$$$\left[\frac{\mathrm{5}−\mathrm{sin}\:^{\mathrm{2}} {X}−\mathrm{sin}\:{X}}{\:\sqrt{{X}}\:+\mathrm{1}}\right]\sqrt{{X}}\: \\ $$$$\frac{\mathrm{5}\sqrt{{X}}}{\:\sqrt{{X}}\:+\mathrm{1}}−\sqrt{{X}}\mathrm{sin}\:{X}\left(\frac{\mathrm{sin}\:{X}\:+\mathrm{1}}{\:\sqrt{{X}}\:+\mathrm{1}}\right) \\ $$$$\: \\ $$$${siut}\:\:{t}=\sqrt{{X}} \\ $$$$\frac{\mathrm{5}{t}}{{t}+\mathrm{1}}−{t}\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)\left(\frac{\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)+\mathrm{1}}{{t}+\mathrm{1}}\right) \\ $$$$ \\ $$$$\frac{\mathrm{5}{t}}{{t}+\mathrm{1}}−\frac{{t}}{{t}+\mathrm{1}}\left[\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)\left(\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)+\mathrm{1}\right]\right. \\ $$$${x}\rightarrow\mathrm{0}+\:\:\:\:\:\:\:{t}\rightarrow+\left[\infty\right. \\ $$$${lim}_{{x}\rightarrow\mathrm{0}+} ={lim}_{{t}\rightarrow+\infty} \frac{\mathrm{5}{t}}{{t}+\mathrm{1}}−\frac{{t}}{{t}+\mathrm{1}}\left[\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)\left(\mathrm{sin}\:\left({t}^{\mathrm{2}} \right)+\mathrm{1}\right]\right. \\ $$$$=\mathrm{5}−{lim}_{{X}\rightarrow+\infty} \left[\mathrm{sin}\left({X}\right)\left(\mathrm{sin}\:\left({X}\right)+\mathrm{1}\right)\right]\: \\ $$$${X}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:\:\:\:\:\mathrm{sin}\:{X}\:\:\rightarrow\mathrm{1} \\ $$$${donc}\:\:{lim}_{{x}\rightarrow\mathrm{0}+} \frac{\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}}\right)}{{x}+\sqrt{{x}}}=\mathrm{5}−\mathrm{2}=\mathrm{3} \\ $$$$ \\ $$
Commented by peter frank last updated on 19/Sep/22
$$\mathrm{font}\:\mathrm{size}\:\mathrm{is}\:\mathrm{too}\:\mathrm{large}.\mathrm{please}\:\mathrm{minimize} \\ $$