Algebraic Techniques for Solving Limits
Finding the limit of a continuous function requires the simple algebraic technique of substitution. For more complicated functions, other algebraic techniques are required. These techniques include...
(The second two options are used when substitution yields a zero in the numerator AND denominator.)
When you have a limit replace the "x" in the problem with the numeric number that x is approaching in the limit.
For example... The limit of (2x-1) when x approaches 2
lim (2x-1)
x->2 ............ we must put substitute 2 for x in the problem
2(2)-1 = ?
3 ......... The answer of this limit is 3.
What is lim (x^2+x-2)/(x^2-1) as x approaches 1?
Substitution cannot be used for this problem because the numerator and denominator will both be zero.
(1+1-2)/(1-1) = 0/0
So we must do it this way... (remember this limit is where x approaches 1)
= lim [(x+2)(x-1)]/[(x+1)(x-1)] ... by factoring
= lim [(x+2)/(x+1)] ... by canceling
= 3/2 ... by substitution
Rationalizing the numerator or denominator
lim [sqr(x) - 2]/(x-4) as x->4
This too yields an indeterminate form through substitution. So we need to do algebra to change the form.
1. rationalize the numerator by multiplying the top and bottom by [sqr(x) + 2]
2. simplify... lim as x approaches 4 of (x-4)/[(x-4)(sqr(x) +2)]
3. cancel factors... lim 1/(sqr(x) +2)
4. substitution... 1/(sqr(4) +2) = 1/4