Question

Consider the following statements :
Statement 1 : \(\lim\limits_{x\rightarrow1}\frac{ax^2+bx+c}{cx^2+bx+a}\) is 1 (where a + b + c ≠ 0)
Statement 2 : \(\lim\limits_{x\rightarrow-2}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\) is \(\frac{1}{4}\)

• A. Only statement 2 is true
• B. Only statement 1 is true
• C. Both statements 1 and 2 are true
• D. Both statements 1 and 2 are false

Answer 1

The Correct Option is B

Let's evaluate both statements:

Statement 1: We are given that \[ \lim_{x \to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a} \] Substituting \( x = 1 \) into the expression, we get \[ \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} = \frac{a + b + c}{c + b + a} = 1 \] Thus, statement 1 is true.

Statement 2: We are given that \[ \lim_{x \to -2} \frac{1}{x + 2} \] As \( x \to -2 \), the denominator approaches 0, making the limit undefined. Therefore, statement 2 is false.

The correct answer is (B) : Only statement 1 is true

Answer 2

Statement 1: \(\lim_{x\rightarrow1}\frac{ax^2+bx+c}{cx^2+bx+a}\)

Substitute \(x = 1\) into the expression:

\(\frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a} = \frac{a+b+c}{c+b+a}\)

Since \(a+b+c \neq 0\), the expression simplifies to:

\(\frac{a+b+c}{a+b+c} = 1\)

Therefore, Statement 1 is true.

Statement 2: \(\lim_{x\rightarrow-2}\frac{\frac{1}{x}+\frac{1}{2}}{x+2}\)

Simplify the expression:

\(\lim_{x\rightarrow-2}\frac{\frac{2+x}{2x}}{x+2} = \lim_{x\rightarrow-2}\frac{x+2}{2x(x+2)}\)

Cancel the \((x+2)\) terms (since \(x \neq -2\)):

\(\lim_{x\rightarrow-2}\frac{1}{2x}\)

Substitute \(x = -2\):

\(\frac{1}{2(-2)} = \frac{1}{-4} = -\frac{1}{4}\)

Therefore, Statement 2 is false, since the limit is \(-\frac{1}{4}\), not \(\frac{1}{4}\).

Conclusion: Only Statement 1 is true.