calculus
posted by Anonymous on .
Evaluate
lim as x> 0
square root (x+1)  square root (2x+1)
/ square root (3x+4)  square root (2x+4)

multiply by [√(3x+4) + √(2x+4)]/[√(3x+4) + √(2x+4)][(√x+1) + √(2x+1)]/[(√x+1) + √(2x+1)]
Expand only the factors which are the difference of squares pattern, leaving the other one alone.
Your limit should reduce to
1([√(3x+4) + √(2x+4)]/[(√x+1) + √(2x+1)]
which as x >
= 1(√4+√4)/(√1+√1)
= 2