Limits, Continuity & Differentiability
Limits at Points of Discontinuity
Grade 12
Question:
<p>Let $f(x) = \frac{|x^3 - 6x^2 + 11x - 6|}{x^3 - 6x^2 + 11x - 6}$, then the number of solutions of $a$, where $\lim_{x \to a} f(x)$ doesn't exist is</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: The limit of an absolute value function divided by itself does not exist where the denominator is zero and the expression changes sign. Factor the polynomial to find these critical points.
<p>First, factor the cubic: $x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$.</p><p>Then $f(x) = \frac{|(x-1)(x-2)(x-3)|}{(x-1)(x-2)(x-3)}$.</p><p>The limit fails to exist at points where the denominator is zero and the sign of the expression changes: $x = 1, 2, 3$.</p><p>At each of these points, the function equals $\pm 1$ on either side with opposite signs.</p>
Correct Answer: C