91Ó°ÊÓ

Answer: To determine the values of \(a\) for which the limit exists and \(\lim_{x\to a} r(x) = r(a)\), we follow the 5-step process. Step 1: Identifying the Rational Function The given rational function is \(r(x) = \frac{x^2 + ax + 1}{x^2 - 4x + 4}\), where \(P(x) = x^2 + ax + 1\) and \(Q(x) = x^2 - 4x + 4\). Step 2: Analyzing the Denominator We analyze the denominator, \(Q(x) = x^2 - 4x + 4\), and look for values of \(x\) that makes it equal to zero. Factoring the denominator, we have \(Q(x) = (x - 2)^2\), so \(Q(x) = 0\) only when \(x = 2\). Step 3: Checking the Limit for a Nonzero Denominator For other values of \(x\), the denominator is nonzero, and we can simply take the limit: \(\lim_{x\to a} r(x) = \lim_{x\to a} \frac{x^2 + ax + 1}{x^2 - 4x + 4}\) Since \(Q(x)\) is nonzero for values of \(x \neq 2\), we can directly substitute \(a\) into the expression: \(\lim_{x\to a} r(x) = \frac{a^2 + a^2 + 1}{a^2 - 4a + 4} = r(a)\) So, for \(a \neq 2\), the limit exists and equals \(r(a)\). Step 4: Checking the Limit for a Zero Denominator Since \(Q(x) = 0\) when \(x = 2\), we need to explicitly check the limit when \(x\) approaches \(2\): \(\lim_{x\to 2} r(x) = \lim_{x\to 2} \frac{x^2 + 2x + 1}{(x - 2)^2}\) As \(x\) approaches \(2\), the numerator approaches \(2^2 + 2(2) + 1 = 9\), and the denominator approaches \((2 - 2)^2 = 0\). The limit does not exist when \(x\) approaches \(2\), so there are no values of \(a\) that satisfy the condition in this case. Step 5: Summarizing the Results The limit of the rational function exists and equals \(r(a)\) for all values of \(a \neq 2\). Therefore, the values of \(a\) for which the limit exists are all real numbers except \(2\).