math
posted by gibbs on .
Evaluate the following limits:
(I) Lim. (x+1)/[1sqrt(4+3x)]
X >1
(II) Lim. (23x5x^2)/(1+2x^2)
X >Infinite

rationalize the denominator
= (x+1)/[1√(4+3x)] * (1 + √(4+3x))/(1 + √(4+3x))
= lim (x+1)(1 + √(4+3x))/ (1  4  3x)
= lim (x+1)(1 + √(4+3x))/ (3(1+x)
= lim (1 + √(4+3x) )/3 , as x >1
= 2/3
Lim. (23x5x^2)/(1+2x^2) , X >Infinite
as x > infinity
the two terms that will "dominate" are 5x^2 on the top and 2x^2 at the bottom, the rest of the terms will become insignificant compared to their value
so we are left with 5x^2/(2x^2 or 5/2
so Lim. (23x5x^2)/(1+2x^2)
X >Infinite
= 5/2