If $a, b$ and $c$ are three distinct real numbers and $\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{2}$, then $a+2 c=$

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Aicharya · Accepted Answer

$\lim _{x ightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{a}$ $=\lim _{x ightarrow \infty} \frac{x^2}{x^2}\left[\frac{(b-c)+(c-a) \frac{1}{x}+\frac{(a-b)}{x^2}}{(a-b)+(b-c) \frac{1}{x}+\frac{(c-a)}{x^2}}ight]=\frac{1}{2}$ $=\frac{b-c}{a-b}=\frac{1}{2}$ $\begin{aligned} & \Rightarrow \quad 2 b-2 c=a-b \ & \Rightarrow \quad 3 b=a+2 c\end{aligned}$

If $a, b$ and $c$ are three distinct real numbers and $\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a)…

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