AP EAMCET · Maths · Limits
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 \(a+2 c=\)
- A \(b\)
- B \(2 b\)
- C \(3 b\)
- D \(4 b\)
Answer & Solution
Correct Answer
(C) \(3 b\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{a}\) \(=\lim _{x \rightarrow \infty} \frac{x^2}{x^2}\left[\frac{(b-c)+(c-a) \frac{1}{x}+\frac{(a-b)}{x^2}}{(a-b)+(b-c) \frac{1}{x}+\frac{(c-a)}{x^2}}\right]=\frac{1}{2}\)…
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