Question

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

Answer 1

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: Factor the quadratic expression. The quadratic expression $x^2 - 5x + 6$ can be factored as: $g(x) = (x-2)(x-3)$

Step 2: Analyze the behavior of $g(x)$ as $x \to 2^{+}$. We are interested in the limit as $x$ approaches $2$ from the right side ($x > 2$). Let $x = 2 + h$, where $h$ is a small positive number ($h \to 0^+$). Substitute $x = 2+h$ into $g(x)$: $g(2+h) = ((2+h)-2)((2+h)-3)$ $g(2+h) = (h)(h-1)$

Now, let's evaluate the limit of $g(x)$ as $h \to 0^+$: $\lim_{h \to 0^+} (h(h-1))$

As $h \to 0^+$: The term $h \to 0^+$. This means $h$ is a very small positive number (e.g., 0.001). The term $(h-1) \to (0-1) = -1$.

So, the product $(h)(h-1)$ will be a very small positive number multiplied by a number very close to $-1$. This results in a very small negative number. For example, if $h=0.001$, then $h(h-1) = (0.001)(-0.999) = -0.000999$. This means that as $x \to 2^+$, $x^2 - 5x + 6$ approaches $0$ from the negative side. We can denote this as $0^-$.

Step 3: Apply the ceiling function. The ceiling function $\lceil y \rceil$ gives the smallest integer greater than or equal to $y$. We need to find $\lceil 0^- \rceil$. Since $0^-$ represents a value that is slightly less than $0$ (e.g., $-0.000999$), the smallest integer greater than or equal to such a value is $0$. For instance: $\lceil -0.000999 \rceil = 0$ $\lceil -0.5 \rceil = 0$ $\lceil -0.0000001 \rceil = 0$

Therefore, $\lim_{x \to 2^{+}}\lceil x^{2}-5x+6 \rceil = \lceil 0^- \rceil = 0$.