Question 228812
For many limits, {{{lim(x->a, f(x)) = f(a)}}} But this does not work with your limit because {{{(sqrt(x)-2)/(x-4)}}} is not defined for x = 4 (because it makes the denominator zero). So we are faced with two options:<ul><li>Use Algebra, including the possible use of properties of limits, to manipulate the expression so that it is defined for x = 4; or</li><li>Decide that there is no limit</li></ul>
The manipulation we will use depends on noticing that each term of the numerator is the square of each term of the denominator. If we notice this and realize that the pattern {{{(a+b)(a-b) = a^2 - b^2}}} can be used to "convert" the numerator to match the current denominator. This will allow us to cancel the (x-4) which is causing the problem. Let's see how this works:
{{{(sqrt(x)-2)/(x-4)}}}
The numerator is of the form (a-b). So multiply the numerator and denominator by it conjugate (a+b), or {{{(sqrt(x)+2)}}}:
{{{((sqrt(x)-2)/(x-4))((sqrt(x) + 2)/(sqrt(x)+2))}}}
{{{((sqrt(x))^2-(2)^2)/((x-4)(sqrt(x)+2))}}}
(If you don't see how the pattern works, resulting in {{{sqrt(x))^2-(2)^2}}}, then use FOIL on the {{{(sqrt(x)-2)*(sqrt(x) + 2)}}} in the numerators.)
Simplifying we get:
{{{(x-4)/((x-4)(sqrt(x)+2)))}}}
When we are finding the limit as x approaches 4, we specifically exclude x=4 as a possible value for x. This is important to understand because if x was actually a 4, then (x-4)/(x-4) would <b>not</b> cancel out. (Zero over zero does not equal one and it does not cancel out.) But since x gets very, very close to but not equal to 4 in this limit, (x-4)/(x-4) is not zero over zero and so it <b>does</b> cancel out. So, after canceling out the (x-4)'s, we have:
{{{1/(sqrt(x)+2)}}}
Now that the (x-4) is gone from the denominator, we have an expression of which we can find the limit:
{{{lim(x->4, ((sqrt(x)-2)/(x-4))) = lim(x->4, (1/(sqrt(x) + 2))) = 1/(sqrt(4) + 2) = 1/(2+2) = 1/4}}}