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

Solution:

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:

a^2-b^2=(a-b)(a+b)

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**