Finding the limit example

Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)

A:  The first thing to do when finding a basic limit is try plugging in the number in question (3).

So, plug in 3 to get:

(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done.  However, if you had just got a number like 4 or something, that would’ve been your answer!

OK, we got 0/0 so that means more work.  More work could mean many things (apply different rules, factor and cancel, simplify, etc).  In this case, it appears we can factor, so we try that:

(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)

Now, here comes some tricky insight.  I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom.  I am going to factor a “-1” out of the (x – 3) that is on top.

Notice:  -1(3 – x) = (x – 3)

So, the numerator becomes:

-1(x – 3)(x2 + 3x + 9)/ (3 – x)

Now, the (x – 3) term cancels from the top and bottom to leave:

-1(x2 + 3x + 9)

So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9).  We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems.  In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:

Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27

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