Question 1092417
<br>The given function can be written as
{{{(x-2)/((x-2)(x-1))}}}<br>
If x is not 2 (so that x-2 is not 0), then we can divide numerator and denominator by (x-2); so everywhere except at x=2, this function is the same as the function
{{{1/(x-1)}}}<br>
That is enough to answer part B of your question.  Except at x=2, the function is everywhere the same as the function 1/(x-1); that means that as you approach 2 from either direction, the function value is getting closer and closer to the value of 1/(x-1) at x=2, which is 1.  Since you approach the same value from either direction, the limit exists at x=2; the limit is 1.<br>
Everywhere except at x=2, the function is the same as 1/(x-1).  To see if the limit exists at x=1, look what happens when you approach x=1 from the left or the right.<br>
If you are approaching from the left, x is less than 1; as you get very close to 1 from the left, (x-1) is negative and gets very close to 0.  That makes the value of 1/(x-1) a large negative number.  So the limit as you approach x=1 from the left is negative infinity.<br>
But when you approach x=1 from the right, (x-1) becomes a very small positive number, so 1/(x-1) becomes a very large positive number.  So the limit as you approach x=1 from the right is positive infinity.<br>
Since the limits as you approach x=1 from the two directions are different, the limit does not exist at x=1.<br>
So the function is discontinuous at x=1 and x=2; in fact, we knew that would be the case in the beginning, since the denominator of the function factors as (x-2)(x-1).<br>
Obviously, since the denominator can't be 0, the points x=1 and x=2 are excluded from the domain; all other values of x are allowed.<br>
We know that the limit as x approaches 2 from either direction is 1.  To see if there are any restrictions on the range, we need to see if there are any x values for which the function value is 1.<br>
{{{1/(x-1) = 1}}}
{{{x-1 = 1}}}
{{{x = 2}}}<br>
The only value of x for which the function value is 1 is x=2, which is not in the domain of the function.  That means there is no value of x for which the y value is 1.  So the range of the function is all real numbers except 1.