We can find the limit by applying L'Hopital's rule:

if f(c) = g(c) = 0, then lim f(x)/g(x), x->c = lim f'(x)/g'(x), x->c

f(x) = x - 3 --> f'(x) = 1 --> f'(3) = 1

g(x) = x^2 - 4x + 3 --> g'(x) = 2x - 4 --> g'(3) = 2

lim f(x)/g(x), x->3 = lim f'(3)/g'(3) = 1/2

To find the limit as x-->3 , ( x-3)/(x^2-4x+3).

Solution

Both numerator and denominators vanish for x=3. Therefore, we should use either the common factor cancelling method or L'Hospital's rule of [f(x)]'/[g(x)]' at x= 3.

Therefore,

limit as x-->3 , ( x-3)/(x^2-4x+3) = (x-3)/[(x-3)(x-1)] at x=3

=1/(x-1) =1/(3-1) =1/2.

2nd method:

Both numerator,x-3 and x^2 - 4x+3 have their zeros at x=3.

So,being qualified under L'Hospital's or Burnoulli's rule,

limit x-->3, (x-3)/(x^2-4x+3) = (x-3)'/(x^2-4x+3)'= 1/(2x-4) at x-3

=1/(2*3-4)

=1/2

As you can see,by substituting the unknown x with the value of 3, we are dealing with a case of indeterminacy, "0/0" type.

lim x->3 ( x-3)/(x^2-4x+3)=(3-3)/(9-12+3)=0/0

It is simple to notice that if "3" cancel the expression from the denominator, that means that 3 is a root of the denominator. We'll find the other root, using Viete's relationships and knowing that:

x1 + x2= -(-4/1)

Bt x1=3, as we've noticed earlier, so:

3+x2=4

x2=4-3

x2=1

Knowing the both roots, now we can write the denominator as follows:

(x^2-4x+3)=(x-x1)(x-x2)

(x^2-4x+3)=(x-3)(x-1)

We'll put back this late expression into the limit:

lim x->3 ( x-3)/(x^2-4x+3)= lim x->3 ( x-3)/(x-3)(x-1)

It is obvious that we'll simplify the common factor (x-3):

lim x->3 ( x-3)/(x-3)(x-1)=lim x->3 1/(x-1)

Now we'll substitute again with the value "3".

lim x->3 1/(x-1)=1/(3-1)=1/2