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Fractions a / b          

The bottom can never equal zero as it zooms off in an undefined way.
The only way for a fraction to be zero is if the top equals zero.

Absolute value ½x ½ This means just take the number part whether it is positive or negative.

Remember that the smallest value of an absolute number is zero unless there are restrictions. As negative and positive absolute values are both positive.½-3½= ½3½ = 3    
Just a warning that these signs have a  different meaning when used with vectors and matrices                                                                      

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Limits

Without too much theory, a limit is the answer an expression goes towards when nearing a certain value. Usually as x tends to a number, the expression that has x  in it, tends towards a certain value.  For a limit to exist it must approach the same number from above and below the value of x.
For example if the limit of a function is 5 as the x - value approaches 20 then the answer must be going towards 5, when x  is 19.9 and again when x  is 20.1.  This is what you mean by a bit above and a bit below the x - value.  19.9 and 20.1 are just numbers that I chose.

Now we will have a problem like we had with domains of functions such as zeroes on the bottom of fractions.  This is a complete no-no!!! Fear not we have ways of making it work!

How to approach limit questions especially when we have zero where we don’t want it:

• Put in the x value and see if it gives an answer that is valid.
• If the answer is invalid, then try to factorise and see if you can cancel out the problem term.
• If factorising doesn’t work or is not an option, then multiply top and bottom by an expression which removes the problem.

So let’s look at examples that will explain when you do it, what you do, and why you do it.

Example One
                     lim_{x® 3} ( x ²  − x  −6 ) / (x  −3)

We can try to put the value 3 in but the bottom of the fraction becomes zero.
Now there is not much we can do with the bottom so let’s look at the top.
Let’s try factorising after all it is only a quadratic.  Remember the two numbers add to give-1 and multiply to give − 6.                Have a quick look at quadratics if you have forgotten

Hint, we hope the bottom of the fraction is one of the brackets, so try a short cut and find a bracket to multiply by (x −3) to give ( x ² − x  −6 ). This time it works and the bracket is (x + 2).
So the limit becomes lim_{x® 3} (x  −3) (x + 2)   = lim_{x® 3} (x + 2) 
                                                           (x −3)

So now we put the value 3 into the limit and we will get lim_{x® 3} (x + 2) = 5.

We must remember to say x ¹ 3.

Example Two
                     lim _{x® 5} ( x ² − 25) /(x  − 5)

We can try to put the value 5 in but the bottom of the fraction becomes zero.
Now there is not much we can do with the bottom again, so let’s look at the top.

We can try factorising, if you look closely the top is the difference of two squares.  This is a common happening so keep a look out for it.  (x ² and 25 (5 ² ). This factorises to the products of the square roots added and subtracted so, ( x ² − 25) = (x  −5)(x + 5).
This is good as we will now be able to cancel the part that was causing the trouble on the bottom of the fraction.

             lim _{x® 5} (x  −5)(x + 5)   =  lim _{x® 5}    (x + 5)
                                   (x  −5)

Now  we can put 5 into the expression so, it goes to a valid answer

             lim _{x® 5}    (x + 5) =  5 + 5 = 10
So the answer is
            lim _{x® 5} ( x ² −25) /(x  −5 )= 10     x ¹ 5

Example Three

lim_{x® 0}    (Ö(x +3) −Ö3 )  
                             x

Please read Ö (x +3) as the square root of (x + 3)

Well, once again we cannot substitute the value in, as the bottom goes to zero.
 Now the top does not look like an easy factorisation.
So what we will do is multiply the top and bottom by an expression which will get rid of our problem usually by cancelling after the multiplication.

Remember the difference of two squares from the last example. 
We are going to multiply by the other bracket.
If the two terms in a bracket are added then we multiply by a bracket with the two terms subtracted 
or
If the two terms in a bracket are subtracted then we multiply by a bracket with the two terms added

         lim_{x® 0}   (Ö(x +3) −Ö3 )  (Ö(x +3) +Ö3 )  
                                       x  (Ö(x +3) +Ö3 )  

So now we have the difference of two squares on top which is written as.

         lim_{x® 0}      ( (x +3) −3 )                       
                           x   (Ö (x +3)   +  Ö 3 )              

            lim_{x® 0}            x                                     
                                 x   (Ö (x +3)   +  Ö 3 ) 

        

          lim_{x® 0}                        1                          
                                      (Ö (x +3)   +  Ö 3 )                                  
 

         ₌                                1            
                               (Ö (0+3)   +  Ö 3 )                                  
 

So the answer is

lim_{x® 0} (Ö(x +3) −Ö3 )   =      1     
                        x                      2 Ö 3

 

What if the limit is tending to infinity, whether positive or negative, what do we do now?
If you think of infinity as being the opposite to zero so just as we wanted to move zero away from the bottom of a fraction we want to put infinity on the bottom of a fraction.  When a big number is on the bottom of a fraction, the answer goes to zero.
So we will divide every term by the highest power of the variable.

 Example Four

                         lim_{x® -¥}         4  −x⁴      
                                                     2x³−5x⁴

So we will divide each term by x⁴.
 

                         lim_{x® -¥}          4/ x⁴ − x⁴/  x⁴   
                                               2x³/  x⁴ −   5x⁴/  x⁴

                        lim_{x® -¥}        4/ x ⁴ −1 
                                                2/  x  −  5                     

Now the values ( 4 / x^{4 )} and ( 2 / x )  become smaller and smaller as x gets bigger and bigger. 
So the limit becomes
                                     lim_{x® -¥}     0  −1        =     −1     =   1    
                                                         0 −5                 −5          5

The answer is

                           lim_{x® -¥}        4 −  x⁴       =    1
                                                  2x³−5x⁴             5

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