Problem: If x is real and positive and grows beyond all bounds, then log3β(6xβ5)βlog3β(2x+1) approaches:
Answer Choices:
A. 0
B. 1
C. 3
D. 4
E. no finite number
Solution:
log3β(6xβ5)βlog3β(2x+1)=log3β2x+16xβ5β=log3β2+x1β6βx5ββ for xξ =0. As x grows beyond all bounds, the last expression approaches log3β26β=log3β3=1.