Q. $\lim_{x\to\infty}\frac{\left(2x^2-3x+5\right)(3x-1)^{x/2}}{\left(3x^2+5x+4\right)\sqrt{(3x+2)^x}}$ is equal to :
A) $\frac2{\sqrt{3e}}$
B) $\frac{2e}{\sqrt3}$
C) $\frac2{3\sqrt e}$
D) $\frac{2e}3$
Explanation
[Solution by Dr. Abubacker] $\frac{2x^2-3x+5}{3x^2+5x+4}\to\frac23$. Also $\frac{(3x-1)^{x/2}}{(3x+2)^{x/2}}=\left(1-\frac3{3x+2}\right)^{x/2}\to e^{-\frac32\cdot\frac13}=e^{-1/2}$. The limit is $\frac2{3\sqrt e}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.