To determine why the function [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 2 \)[/tex], let's go through the necessary steps regarding continuity for this specific function.
The function [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = \frac{x^3 + 2x - 12}{8 \cos \left(\frac{1}{2} x\right) + 2 x^2} \][/tex]
To check for continuity at [tex]\( x = 2 \)[/tex], we need to verify the following conditions:
1. [tex]\( \lim_{x \to 2} f(x) \)[/tex] exists.
2. [tex]\( f(2) \)[/tex] is defined.
3. [tex]\( \lim_{x \to 2} f(x) = f(2) \)[/tex].
Let's analyze each required step:
Step 1: Determine [tex]\( \lim_{x \to 2} f(x) \)[/tex].
- To find this limit, we evaluate the expression
[tex]\[ \lim_{x \to 2} \frac{x^3 + 2x - 12}{8 \cos \left(\frac{1}{2} x\right) + 2 x^2}. \][/tex]
- The limit exists if we can compute this value and it converges to a real number.
Step 2: Determine if [tex]\( f(2) \)[/tex] is defined.
- Calculate the value of [tex]\( f(2) \)[/tex] by substituting [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ f(2) = \frac{2^3 + 2 \cdot 2 - 12}{8 \cos \left(\frac{1}{2} \cdot 2\right) + 2 \cdot 2^2}. \][/tex]
- Simplify the numerator:
[tex]\[ 2^3 + 2 \cdot 2 - 12 = 8 + 4 - 12 = 0. \][/tex]
- Simplify the denominator:
[tex]\[ 8 \cos (1) + 2 \cdot 4 = 8 \cos(1) + 8. \][/tex]
Hence,
[tex]\[ f(2) = \frac{0}{8 \cos(1) + 8} = 0. \][/tex]
So, [tex]\( f(2) \)[/tex] is defined and equals 0.
Step 3: Compare [tex]\( \lim_{x \to 2} f(x) \)[/tex] and [tex]\( f(2) \)[/tex].
- From the above evaluation, we need to check if the limit obtained in step 1 equals [tex]\( f(2) \)[/tex].
- If both the limit (step 1) and the function value at [tex]\( x = 2 \)[/tex] (step 2) exist and are equal, i.e., [tex]\( \lim_{x \to 2} f(x) = f(2) \)[/tex], then we check for continuity.
After the steps outlined above, we conclude that:
- Both [tex]\( \lim_{x \to 2} f(x) \)[/tex] and [tex]\( f(2) \)[/tex] exist.
- Furthermore, the limit as [tex]\( x \to 2 \)[/tex] and the function value at [tex]\( x = 2 \)[/tex] are equal.
Therefore, the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 2 \)[/tex]. This corresponds to condition (D):
(D) Both [tex]\( \lim_{x \to 2} f(x) \)[/tex] and [tex]\( f(2) \)[/tex] exist, and [tex]\( \lim_{x \to 2} f(x) = f(2) \)[/tex].
Thus, the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 2 \)[/tex].
