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Question Number 193874 by cortano12 last updated on 22/Jun/23

Answered by Subhi last updated on 22/Jun/23

$$$$$$cos\left(y\right)=1-2{sin}^2\left(\frac{y}{2}\right)$$$$1-cos({cx}^2+bx+a)=2{sin}^2\left(\frac{{cx}^2+bx+a}{2}\right)$$$${lim}_{x\to \frac{1}{\alpha }}\frac{2{sin}^2\left(\frac{{cx}^2+bx+a}{2}\right)}{{(1-\alpha x)}^2}$$$${ax}^2+bx+c\Rrightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$$$suppose:\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a},\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$$$$({ax}^2+bx+c)\Rrightarrow \alpha \beta =\frac{c}{a}$$$$({cx}^2+bx+a)\Rrightarrow \delta \gamma =\frac{a}{c}$$$$\frac{a}{c}={\left(\frac{c}{a}\right)}^{-1}$$$$\therefore \frac{1}{\alpha }=\frac{1}{\delta },\frac{1}{\beta }=\frac{1}{\gamma }$$$$({ax}^2+bx+c)=a(x-\alpha )(x-\beta )$$$$({cx}^2+bx+a)=c(x-\frac{1}{\alpha })(x-\frac{1}{\beta })=\frac{c}{\alpha \beta }(1-\alpha x)(1-\beta x)=a(1-\alpha x)(1-\beta x)$$$${lim}_{x\to \frac{1}{\alpha }}\frac{2{sin}^2(a\left(\frac{1-\beta x)}{2}\right).(1-\alpha x))}{{(1-\alpha x)}^2}=\frac{2a^2{(1-\beta x)}^2}{4}=\frac{a^2{(1-\beta x)}^2}{2}=\frac{a^2{(1-\frac{\beta }{\alpha })}^2}{2}$$$$\frac{a^2{(1-\frac{\beta }{\alpha })}^2}{2}=\frac{a^2{(1-\frac{-b-\sqrt{b^2-4ac}}{-b+\sqrt{b^2-4ac}})}^2}{2}=\frac{2a^2(b^2-4ac)}{{(-b+\sqrt{b^2-4ac})}^2}=\frac{(b^2-4ac)}{2{\alpha }^2}$$

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