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: Divide by x^30
Dividing top and bottom by $x^{30}$ gives $\dfrac{(2 - 1/x)^{19}(3 + 2/x)^{11}}{(6 - 5/x)^{30}}$.

Step 2: Limit
As $x \to \infty$ this is $2^{19}3^{11}/6^{30} = 2^{-11}3^{-19}$.

Step 3: Add exponents
$a + b = -11 - 19 = -30$. Option (A).

Final Answer:
Option (A). \[ \boxed{-30} \]