On-line Math 21

1.3  Infinite Limits

Example 4

lim x® ¥  x2+5x-6 x3+x+5

Hints

For this one, we still can make it clearer what happens to the dominant terms by dividing out the top and bottom by the largest shared power of x .

More

In this case, the largest shared power is x² . Dividing out top and bottom gives:

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

Finish it

Now, as x goes to infinity, all the terms of either the numerator or denominator with powers of x in their denominator do go to 0, but there is that one term of x in the denominator which goes to infinity, bringing the whole denominator off to infinity with it, and so the limit goes to 0,

lim x® ¥  x2+5x-6 x3+x+5 = 0.

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Copyright (c) 2000 by David L. Johnson.