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« Properties of Limits |

Limits: (lesson 2 of 5)

Example 1: Find the limit

Solution

we will use :

Example 2:

Solution :

Direct substitution gives the indeterminate form 0/0. The numerator can be separated into the product of the two binomials (x+5) and (x-2).

So the limit is equivalent to

From here, we can simply divide (x - 2) out of the fraction. We do not have to worry about (x - 2) being equal to 0, since in the context of this limit, the expression can be treated as if x will never equal 2.

This gives us The expression inside the limit is now linear, so the limit can be found by direct substitution. This obtains 2 + 5 = 7.

We then can say that

Example 3:

Solution:

Example 4:

Find the limit

Solution:

Example 5: