Let’s take the limit of another rational function we’ll get the limit as x approaches infinity of 2x³ minus 9x² minus 2x minus 9 over 40x² plus 8x³. Be careful here. I’ve written the bottom, the bottom polynomial with the terms out of order so when you’re using the trick of multiplying by 1 over the highest power of x in the denominator, make sure you check for the highest power of x, it’s actually x³ here.
’m going to multiply by 1 over x³ and 1 over x³ and let’s see what that gives me. I get the limit as x approaches infinity of 2x³ times 1 over x³ is 2, 9x² times 1 over x³ is 9 over x minus 2x times 1 over x² is minus 2, I’m sorry all over x³ is minus 2 over x² and I get minus 9 over x³. And then the denominator we get 40x² times 1 over x³, that’s 40 over x, and then plus 8x³ times 1 over x³, x plus 8.
What happens? Each of these guys is going to go to zero as x goes to infinity. As the denominators get big, these guys get small. And so does this guy going to zero. Now what’s left is 2 plus a bunch of zero and zero plus 8. So the final limit is 2/8 or ¼.
That’s it, always remember the trick of multiplying by the reciprocal of the highest power of x in the denominator and remember later on when you’re calculating more limits in calculus, this trick only for limit of x approaches infinity or as x approaches negative infinity.