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Question Number 111083 by bemath last updated on 02/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\left(1\right)\int \frac{\mathrm{dx}}{3\sin \mathrm{x}+\sin {}^3\mathrm{x}}$$$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }\mathrm{x}(5^{\frac{1}{\mathrm{x}}}-1)$$$$\left(3\right)\mathrm{find}\mathrm{the}\mathrm{asymptotes}\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$$

Answered by john santu last updated on 02/Sep/20

$$\left(2\right)\underset{x\to \mathrm{\infty }}{\lim }x(5^{\frac{1}{x}}-1)$$$$let\frac{1}{x}=t;t\to 0$$$$\underset{t\to 0}{\lim }\frac{5^t-1}{t}=\underset{t\to 0}{\lim }\frac{\ln \left(5\right).5^t}{1}=\ln \left(5\right)$$

Answered by john santu last updated on 02/Sep/20

$$\left(3\right)lettheasymptotesisy=mx+c$$$$wherem=\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{y}{x}\right)$$$$m=\underset{x\to \mathrm{\infty }}{\lim }\frac{\pm b\sqrt{x^2-a^2}}{ax}=\frac{\pm b}{a}\underset{x\to \mathrm{\infty }}{\lim }\sqrt{1-\frac{a^2}{x^2}}=\pm \frac{b}{a}$$$$andc=\underset{x\to \mathrm{\infty }}{\lim }(y-mx)=\underset{x\to \mathrm{\infty }}{\lim }[\pm \frac{b}{a}\sqrt{x^2-a^2}-(\pm \frac{bx}{a})]$$$$=\pm \frac{b}{a}\underset{x\to \mathrm{\infty }}{\lim }(\sqrt{x^2-a^2}-x)$$$$=\pm \frac{b}{a}\underset{x\to \mathrm{\infty }}{\lim }\frac{-a^2}{\sqrt{x^2-a^2}+x}=0$$$$Hence,theasymptotesof$$$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1isy=\pm \frac{b}{a}x$$

Commented by bobhans last updated on 02/Sep/20

$$\mathrm{thank}\mathrm{you}$$

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