Question:
If \(a,b,c\) are three distinct real numbers and
\[
\lim_{x\to \infty}
\frac{(b-c)x^2+(c-a)x+(a-b)}
{(a-b)x^2+(b-c)x+(c-a)}
=
\frac{1}{2}
\]
then \(a+2c=\)
If \(a,b,c\) are three distinct real numbers and
\[
\lim_{x\to \infty}
\frac{(b-c)x^2+(c-a)x+(a-b)}
{(a-b)x^2+(b-c)x+(c-a)}
=
\frac{1}{2}
\]
then \(a+2c=\)
Show Hint
For limits of rational functions as \(x\to\infty\), compare the highest powers of \(x\). If numerator and denominator have the same degree, the limit is the ratio of their leading coefficients.
Updated On: Jun 22, 2026
- \(b\)
- \(2b\)
- \(3b\)
- \(4b\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Identify the highest degree terms.
The given limit is \[ \lim_{x\to \infty} \frac{(b-c)x^2+(c-a)x+(a-b)} {(a-b)x^2+(b-c)x+(c-a)} = \frac{1}{2} \] Since numerator and denominator are both quadratic polynomials, the limit as \(x\to\infty\) is the ratio of coefficients of \(x^2\).
So, \[ \frac{b-c}{a-b}=\frac{1}{2} \]
Step 2: Cross multiply.
\[ 2(b-c)=a-b \] \[ 2b-2c=a-b \]
Step 3: Rearrange the equation.
Bring all terms involving \(a,b,c\) together: \[ a-b=2b-2c \] \[ a=3b-2c \] Therefore, \[ a+2c=3b \]
Step 4: Final conclusion.
Hence, \[ \boxed{3b} \]
The given limit is \[ \lim_{x\to \infty} \frac{(b-c)x^2+(c-a)x+(a-b)} {(a-b)x^2+(b-c)x+(c-a)} = \frac{1}{2} \] Since numerator and denominator are both quadratic polynomials, the limit as \(x\to\infty\) is the ratio of coefficients of \(x^2\).
So, \[ \frac{b-c}{a-b}=\frac{1}{2} \]
Step 2: Cross multiply.
\[ 2(b-c)=a-b \] \[ 2b-2c=a-b \]
Step 3: Rearrange the equation.
Bring all terms involving \(a,b,c\) together: \[ a-b=2b-2c \] \[ a=3b-2c \] Therefore, \[ a+2c=3b \]
Step 4: Final conclusion.
Hence, \[ \boxed{3b} \]
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