Question

If \(\underset{x\rightarrow \infty }{lim}\frac{(2x-1)^{19}\cdot (3x+2)^{11}}{(6x-5)^{30}} = 2^a\cdot 3^b\), then \(a+b =\)

• A. \(-30\)
• B. \(-11\)
• C. \(-19\)
• D. \(30\)

Hint:

Compare leading coefficients since the degrees match: 19 + 11 = 30.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
When numerator and denominator have the same degree, the limit at infinity is the ratio of leading coefficients.

Step 2: Compute
The numerator has degree \(19 + 11 = 30\) and the denominator has degree 30.
\[ \lim_{x\to\infty} \frac{(2x - 1)^{19}(3x + 2)^{11}}{(6x - 5)^{30}} = \frac{2^{19}\cdot 3^{11}}{6^{30}} \]
\[ = \frac{2^{19} 3^{11}}{2^{30} 3^{30}} = 2^{-11}\cdot 3^{-19} \]

Step 3: Read off
So \(a = -11\) and \(b = -19\), giving \(a + b = -30\). Option (B) is only a, and (C) is only b.

Final Answer:
The value of \(a + b\) is \(-30\), option (A). \[ \boxed{-30} \]