If lim_{x to infty} frac{(2x - 1)^{19} cdot ( | vedclass

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(A) To evaluate the limit as $x 	o \infty$, we consider the highest power of $x$ in the numerator and denominator.Numerator: $(2x - 1)^{19} \cdot (3x

Vedclass · Accepted Answer

$-30$

Vedclass · Answer

(A) To evaluate the limit as $x 	o \infty$, we consider the highest power of $x$ in the numerator and denominator.Numerator: $(2x - 1)^{19} \cdot (3x + 2)^{11} \approx (2x)^{19} \cdot (3x)^{11} = 2^{19} \cdot 3^{11} \cdot x^{30}$.Denominator: $(6x - 5)^{30} \approx (6x)^{30} = 6^{30} \cdot x^{30}$.Thus, the limit is $\frac{2^{19} \cdot 3^{11}}{6^{30}} = \frac{2^{19} \cdot 3^{11}}{(2 \cdot 3)^{30}} = \frac{2^{19} \cdot 3^{11}}{2^{30} \cdot 3^{30}}$.Simplifying the powers: $2^{19-30} \cdot 3^{11-30} = 2^{-11} \cdot 3^{-19}$.Comparing this with $2^a \cdot 3^b$, we get $a = -11$ and $b = -19$.Therefore, $a + b = -11 + (-19) = -30$.

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

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