Question

Consider the two functions \(f_1(x) = \dfrac{x^2-4}{x-2}\) and \(f_2(x) = x^2 - 2x + 2\). Which of the following is the value of \((f_1(x)+f_2(x))\) as \(x \to 2\)?

• A. 0
• B. 6
• C. 8
• D. \(\infty\)

Hint:

Factor x^2-4=(x-2)(x+2) to cancel the removable discontinuity in f1(x), then just substitute x=2 into f2(x).

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
We have \(f_1(x) = \dfrac{x^2-4}{x-2}\) and \(f_2(x) = x^2 - 2x + 2\). We need the value of \(f_1(x) + f_2(x)\) as \(x \to 2\).

Step 2: Key Formula or Approach.
For a sum of two functions, the limit of the sum equals the sum of the limits, provided each limit exists:
\[ \lim_{x \to 2} \left[f_1(x) + f_2(x)\right] = \lim_{x \to 2} f_1(x) + \lim_{x \to 2} f_2(x) \]
For \(f_1(x)\), notice \(x^2 - 4 = (x-2)(x+2)\), so the \((x-2)\) factor cancels with the denominator for \(x \ne 2\), removing the removable discontinuity at \(x=2\).

Step 3: Detailed Explanation.
Simplify \(f_1(x)\) for \(x \ne 2\):
\[ f_1(x) = \frac{(x-2)(x+2)}{x-2} = x+2 \]
So \(\lim_{x\to2} f_1(x) = 2+2 = 4\).
Now evaluate \(f_2(x)\) at \(x=2\) directly, since it is a polynomial and has no discontinuity anywhere:
\[ f_2(2) = (2)^2 - 2(2) + 2 = 4 - 4 + 2 = 2 \]
So \(\lim_{x\to2} f_2(x) = 2\).
Add the two limits:
\[ \lim_{x\to2}\left[f_1(x)+f_2(x)\right] = 4 + 2 = 6 \]

Step 4: Final Answer.
The value is 6, which is option (B). Options (A) 0 and (D) infinity ignore that \(f_1\) has a finite removable-discontinuity limit of 4, not 0 or unbounded; option (C) 8 comes from wrongly evaluating \(f_2(2)\) as 4 instead of the correct value 2. \[ \boxed{6} \]