document.write( "Question 1092417: A. Find the limit of f(x) = x - 2 / x^2 - 3x + 2 as x approaches 1, or explain why the limit does not exist.
\n" ); document.write( " B. Find the limit of f(x) = x - 2 / x^2 - 3x + 2 as x approaches 2, or explain why the limit does not exist.
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\n" ); document.write( "C. Apply the definition of continuity to identify any points of discontinuity in the function f, showing all work. 1. Explain the effect of the discontinuities identified in part C on the domain and range of the function f. \n" ); document.write( "
Algebra.Com's Answer #707017 by greenestamps(13214)\"\" \"About 
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The given function can be written as
\n" ); document.write( "\"%28x-2%29%2F%28%28x-2%29%28x-1%29%29\"

\n" ); document.write( "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
\n" ); document.write( "\"1%2F%28x-1%29\"

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "Since the limits as you approach x=1 from the two directions are different, the limit does not exist at x=1.

\n" ); document.write( "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).

\n" ); document.write( "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.

\n" ); document.write( "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.

\n" ); document.write( "\"1%2F%28x-1%29+=+1\"
\n" ); document.write( "\"x-1+=+1\"
\n" ); document.write( "\"x+=+2\"

\n" ); document.write( "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.
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