Menu Close

1-lim-x-1-x-2-1-3-sin-1-x-1-3-2-lim-x-x-2-1-sin-2-x-x-sin-x-2-

Tinku Tara 0 Comments
Pin




Question Number 122130 by bemath last updated on 14/Nov/20
(1) lim_(x→1) (x^2 −1)^3 sin ((1/(x−1)))^3 ? (2) lim_(x→∞) ((x^2 (1+sin^2 x))/((x+sin x)^2 )) ?
$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{3}} ? \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)}{\left({x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:? \\ $$
Answered by TANMAY PANACEA last updated on 14/Nov/20
1)1≥sin((1/(x−1)))≥−1 so lim_(x→1) (x^2 −1)^3 sin((1/(x−1)))^3 =0×(any value in between ±1)=0
$$\left.\mathrm{1}\right)\mathrm{1}\geqslant{sin}\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)\geqslant−\mathrm{1} \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} {sin}\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{3}} \\ $$$$=\mathrm{0}×\left({any}\:{value}\:{in}\:{between}\:\pm\mathrm{1}\right)=\mathrm{0} \\ $$
Answered by Bird last updated on 14/Nov/20
2)f(x)=((x^2 (1+sin^2 x))/((x+sinx)^2 )) ⇒ f(x)=((x^2 +x^2 sin^2 x)/(x^2 +2xsinx +sin^2 x)) =((x^2 (1+sin^2 x))/(x^2 (1+((2sinx)/x)+((sin^2 x)/x^2 )))) =((1+sin^2 x)/(1+((2sinx)/x)+((sin^2 x)/x^2 ))) ⇒ lim_(x→∞) f(x)=lim_(x→∞) 1+sin^2 x =lim_(x→+∞) 1+((1−cos2x)/2) =lim_(x→∞) (1/2){3−cos(2x)} but cos(2x) havent any limit at∞ ⇒lim f(x) dont exist !
$$\left.\mathrm{2}\right){f}\left({x}\right)=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right)}{\left({x}+{sinx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} \:+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xsinx}\:+{sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}{sinx}}{{x}}+\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{\mathrm{1}+\frac{\mathrm{2}{sinx}}{{x}}+\frac{{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow\infty} \mathrm{1}+{sin}^{\mathrm{2}} {x} \\ $$$$={lim}_{{x}\rightarrow+\infty} \mathrm{1}+\frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}} \\ $$$$={lim}_{{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{3}−{cos}\left(\mathrm{2}{x}\right)\right\} \\ $$$${but}\:{cos}\left(\mathrm{2}{x}\right)\:{havent}\:{any}\:{limit}\:{at}\infty \\ $$$$\Rightarrow{lim}\:{f}\left({x}\right)\:{dont}\:{exist}\:\:! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *