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

A different way to handle the $\frac{0}{0}$ form in $f_1(x)$ at $x=2$ is to check it approaches a finite number using L'Hopital's rule, then add in $f_2$ separately.

1. Check the form of $f_1(x)$ at $x=2$: plugging $x=2$ directly into $\dfrac{x^2-4}{x-2}$ gives $\dfrac{0}{0}$, an indeterminate form, so direct substitution alone does not work and we need another route.
2. Apply L'Hopital's rule: differentiate the numerator and denominator of $f_1(x)$ separately with respect to $x$: numerator derivative is $2x$, denominator derivative is $1$. So
   $$\lim_{x\to2} f_1(x) = \lim_{x\to2}\frac{2x}{1} = 2(2) = 4$$
3. Evaluate $f_2(x)$ by direct substitution: $f_2(x) = x^2-2x+2$ is a polynomial, so it is continuous everywhere and its limit at $x=2$ equals its value there:
   $$f_2(2) = 4 - 4 + 2 = 2$$
4. Add the two results: $$\lim_{x\to2}\left[f_1(x)+f_2(x)\right] = 4+2 = 6$$
5. Why the distractors are wrong: option (A) 0 would only be right if $f_1$ vanished at $x=2$, but it approaches 4, not 0. Option (D) infinity would apply if the denominator did not cancel, but the factor $(x-2)$ cancels cleanly, so there is no blow-up. Option (C) 8 comes from a slip in evaluating $f_2(2)$, misreading it as 4 instead of 2.

Let's summarize:

• $f_1(x)$ has a removable discontinuity at $x=2$; its limit there is $4$, found either by factoring or L'Hopital's rule.
• $f_2(x)$ is a polynomial, so its limit at $x=2$ is just $f_2(2) = 2$.

Adding the two limits gives $4+2=6$, so the answer is option (B).