Question Number 111083 by bemath last updated on 02/Sep/20
$$\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\left(\mathrm{1}\right)\int\:\frac{\mathrm{dx}}{\mathrm{3sin}\:\mathrm{x}+\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{x}}} \:−\mathrm{1}\right)\: \\ $$$$\left(\mathrm{3}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{asymptotes}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1}\: \\ $$
Answered by john santu last updated on 02/Sep/20
$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\:\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} \:−\:\mathrm{1}\right) \\ $$$${let}\:\frac{\mathrm{1}}{{x}}\:=\:{t}\:;\:{t}\rightarrow\mathrm{0} \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}^{{t}} −\mathrm{1}}{{t}}\:=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{5}\right).\mathrm{5}^{{t}} }{\mathrm{1}}\:=\:\mathrm{ln}\:\left(\mathrm{5}\right) \\ $$
Answered by john santu last updated on 02/Sep/20
![(3) let the asymptotes is y = mx+c where m = lim_(x→∞) ((y/x)) m=lim_(x→∞) ((±b (√(x^2 −a^2 )))/(ax)) = ((±b)/a)lim_(x→∞) (√(1−(a^2 /x^2 ))) = ± (b/a) and c = lim_(x→∞) (y−mx) = lim_(x→∞) [±(b/a)(√(x^2 −a^2 ))−(±((bx)/a)) ] = ±(b/a)lim_(x→∞) ((√(x^2 −a^2 )) −x ) = ±(b/a) lim_(x→∞) ((−a^2 )/( (√(x^2 −a^2 ))+x)) = 0 Hence, the asymptotes of (x^2 /a^2 ) − (y^2 /b^2 ) = 1 is y = ±(b/a)x](https://www.tinkutara.com/question/Q111088.png)
$$\left(\mathrm{3}\right)\:{let}\:{the}\:{asymptotes}\:{is}\:{y}\:=\:{mx}+{c}\: \\ $$$${where}\:{m}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{y}}{{x}}\right) \\ $$$${m}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\pm{b}\:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{ax}}\:=\:\frac{\pm{b}}{{a}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\:=\:\pm\:\frac{{b}}{{a}} \\ $$$${and}\:{c}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({y}−{mx}\right)\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\pm\frac{{b}}{{a}}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\left(\pm\frac{{bx}}{{a}}\right)\:\right] \\ $$$$=\:\pm\frac{{b}}{{a}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:−{x}\:\right) \\ $$$$=\:\pm\frac{{b}}{{a}}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{−{a}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }+{x}}\:=\:\mathrm{0} \\ $$$${Hence},\:{the}\:{asymptotes}\:{of}\: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:−\:\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:{is}\:{y}\:=\:\pm\frac{{b}}{{a}}{x} \\ $$
Commented by bobhans last updated on 02/Sep/20
$$\mathrm{thank}\:\mathrm{you} \\ $$