Expert: Scotto - 4/7/2009

Question
lim     ((x-2)^7)(-2x^2)
x->1


So for this, I got a limit of 2, but I need to state all the limit laws I used.. I don't really know in this problem how to find 2 using limits. I just plugged it in. So how do I know which limit laws to use first. Could you run me through the process and which laws to use? Thanks.  

Answer
The lim as x->1 of x-2 is -1.  
You can do this by direct substitution.
When -1 is taken to the 7th, the answer is still -1.

the lim as x->1 of -2x^2 is -2.
This, too, can be done by direct substitution.
The value of x, when squared, is 1.  
When multiplied by -2, the answer is -2.

When the entire fraction is looked at, the result is -1/-2,
which is that same as 1/2.

The only time you can't do direct substitution is when 0 or infinity is gotten in the numerator or the denominator.

An example of this would be lim x->2 of (x^2-4x+4)/(x-2).
Putting in the value of x gives 0/0.  
This means that x can never be 0.
However, the limit can be found.
The top of this fraction is (x-2)^2 and the bottom is (x-2).
One of the (x-2)'s in the top and the (x-2) in the bottom can cancel,
leaving us with the lim x->2 of x-2.
Now where 2 is put in, it can be seen the limit really is 0.

Another example of this would be lim x->2 of (x^2-5x+6)/(x-2).
Putting in the value of x gives 0/0.  
The top of this fraction is (x-2)(x-3) and the bottom is (x-2).
The (x-2) in the top and the (x-2) in the bottom can cancel,
leaving us with the lim x->2 of x-3.
Now where 2 is put in, it can be seen the limit really is -1.