On-line Math 21

1.3  Infinite Limits

Example 2

lim x® ¥  2x3+x2+x-6 x3+4x+5

Hints

The thing to notice here is that both the numerator and the denominator are headed to infinity as x gets larger. But, they are headed to infinity at about the same speed. Yes, the numerator will be a bit larger than the denominator, but not significantly so.

Try to change the expression (while keeping the values the same...) so that the significant terms are more isolated.

Be More Specific.

The way to emphasize the most significant terms in the numerator and denominator is to divide out the top and bottom by the largest common power of x , in this case x³ :

lim x® ¥  2x3+x2+x-6 x3+4x+5 = lim x® ¥  ( 2x3+x2+x-6) 1 x3 ( x3+4x+5) 1 x3 = lim x® ¥  2+ 1 x + 1 x2 - 6 x3 1+ 4 x2 + 5 x3 .

Finish the solution.

From here, all you have to do is notice that

lim x® ¥  1 x = 0, lim x® ¥  1 x2 = 0, and lim x® ¥  1 x3 = 0.

The only terms left standing are then the dominant ones, the 2 in the numerator and the 1 in the denominator. The final answer is 2.

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Copyright (c) 2000 by David L. Johnson.