Question

$\lim_{x \to 2^+} \lceil x^2 - 5x + 6 \rceil$

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

0

Answer 2

To evaluate the limit $\lim_{x \to 2^+} \lceil x^2 - 5x + 6 \rceil$, we first analyze the expression inside the ceiling function, let $g(x) = x^2 - 5x + 6$.

Step 1: Evaluate the inner function at the limit point. Substitute $x=2$ into $g(x)$: $g(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$.

Step 2: Determine the behavior of the inner function as $x$ approaches $2$ from the right side ($x \to 2^+$). We need to know if $g(x)$ approaches $0$ from values greater than $0$ (i.e., $0^+$) or from values less than $0$ (i.e., $0^-$). The quadratic expression $g(x) = x^2 - 5x + 6$ can be factored as $g(x) = (x-2)(x-3)$.

As $x \to 2^+$:

• The term $(x-2)$ will be a small positive number (e.g., if $x=2.001$, $x-2 = 0.001$). We denote this as $0^+$.
• The term $(x-3)$ will approach $2-3 = -1$.

Therefore, $g(x) = (x-2)(x-3)$ approaches $(0^+)(-1)$. The product of a small positive number and a negative number is a small negative number. So, $g(x)$ approaches $0$ from the negative side, which we denote as $0^-$.

Step 3: Evaluate the ceiling function with the determined behavior. Let $y = g(x)$. As $x \to 2^+$, $y \to 0^-$. Now we need to find $\lim_{y \to 0^-} \lceil y \rceil$. The ceiling function $\lceil y \rceil$ gives the smallest integer greater than or equal to $y$. If $y$ is a number slightly less than $0$ (for example, $y = -0.001$, $y = -0.5$, or $y = -0.99$), the smallest integer greater than or equal to $y$ is $0$. For instance: $\lceil -0.001 \rceil = 0$ $\lceil -0.5 \rceil = 0$ $\lceil -0.99 \rceil = 0$

Thus, $\lim_{y \to 0^-} \lceil y \rceil = 0$.

Combining these steps, we get: $\lim_{x \to 2^+} \lceil x^2 - 5x + 6 \rceil = \lim_{y \to 0^-} \lceil y \rceil = 0$.