Lim (4x^3-3x^2-5)/(6x^3+4x+1) as x--infinity

Question

find the limits

Answer 1

As x gets larger the constants become insignificant so the expression becomes:

(4x³-3x²)/(6x³+4x)=(4x²-3x)/(6x²+4). Again, the constant becomes negligible:

(4x²-3x)/6x²=(4x-3)/6x. Dropping the constant again: 4x/6x=⅔, which is the limit as x→∞.

Actually, we can ignore all but the highest degree of x and we get the same result.