Box "indeterminate difference"

Here we will address the following problem: We want to find a limit of a sequence and after trying to substitute infinity, we obtained the indeterminate expression ∞ − ∞.

There is a standard procedure, but it leads to rather ugly expressions so we will leave it as the last resort and first expore some alternatives:

1. If the two infinities involve powers, exponentials, logarithms, factorials, roots and their combinations, and the dominant powers are different, we can find the answer by factoring, see the box "polynomials and ratios with powers".

2. If the two infinities are actually roots, we can often fix the problem by getting rid of the roots and some cancelling, see the box "difference of roots".

3. Sometimes the two "infinities" allow natural algebraic simplification into one fraction, typically in the from of some common denominator. Such ratios are very often of the indeterminate type and we can use the l'Hospital rule.

4. Sometimes we can factor out some factor and get a product. If we are lucky, this product can be evaluated right away, often we would get a product of indeterminate form and we would solve it using the appropriate trick, see the box "indeterminate product".

In general, we can express it as follows: A − B = A⋅(1 − B/A). If the ratio B/A goes to 1, we get the indeterminate type ∞⋅0, otherwise we get the answer right away.

Example: The expression n³ − n² is of the type ∞ − ∞. We try to factor out the first term and obtain

n³ − n² = n³(1 − 1/n).

When we try to put in infinity, we get ∞³⋅(1 − 1/∞) = ∞⋅(1 − 0) = ∞. In fact, note that we did exactly the factoring business from the first alternative, since n³ is the dominant power.

5. We can always transform the difference into a fraction as follows:

What is the type of this new expression?

This is an indeterminate ratio and we can solve it using l'Hospital's rule. However, as you can see, the resulting fraction is quite complicated, so the calculations tend to be long and it is better to first try one of the previous tricks and use this one only as the last resort.

Just like the l'Hospital rule, this universal transformation also sometimes fails. We will show how it fares when applied to the example above:

As you can see, it did not help.

In Solved Problems - Limits, these methods are used in this problem.

Next box: indeterminate power
Back to Methods Survey - Limits