lim (x-2)/(x^2-4), x->2
first we find the function value when x = 2
then (2-2)/(4-4) = 0/0
0/0 means that the substitution methid failed becase both upper and lower functions has a the same root (2)
Now we try to factor the function to try and eleminate the common factor
==> (x-2)/(x-2)(x+2) = 1/(x+2)
then lim 1/(x+2) when x--> 2 is 1/(2+2) = 1/4
then lim (x-2)/(x^2-4), x->2 equals 1/4
To find Lt (x-2)/(x^2-4) as x--> 2
Let f(x) = (x-2)/(x^22-4).
Then f(2) = 0/0 form, which is indeterminate.Such limits could be taken after dividing by the common factor of x-2. Or using the L'Hospital's rule.
Lt x--> 2 f(x) = lt x-->2 (x-2)/(x^2-4) = Lt x-->2 (x-2)/[(x-2)(x+2)] =lt x-->2 [1/(x+2)] = 1/(2+2) = 4.
To calculate the limit, we'll apply the dividing out technique.
We'll apply the direct substitution, by substituting the unknown x, by the value 2 and we'll see that it fails, because both, numerator and denominator, are cancelling for x=2. That means x=2 is a root for both, that means that (x-2) is a common factor for both.
We'll write the denominator using the formula:
x^2-4 = (x-2)(x+2)
Now, we'll evaluate the limit:
lim (x-2)/(x^2-4)= lim (x-2)/(x-2)(x+2)
Now, we can divide out like factor:
lim (x-2)/(x^2-4) = lim 1/(x+2)
We can apply the replacement theorem and we'll get:
lim 1/(x+2) = 1/(2+2) = 1/4
So, lim (x-2)/(x^2-4) = 1/4