Answer :
Certainly! Let's derive the value of [tex]\( A \)[/tex] for the given function [tex]\( f(x) \)[/tex] so that it is continuous at [tex]\( x = 0 \)[/tex]. To ensure continuity, we need the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 to equal [tex]\( f(0) \)[/tex], which is [tex]\( A \)[/tex].
Given the function:
[tex]\[ f(x) = \begin{cases} \frac{1 - \cos 6x}{4x^2} & \text{if } x \neq 0 \\ A & \text{if } x = 0 \end{cases} \][/tex]
To find [tex]\( A \)[/tex], we evaluate the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} \][/tex]
To determine this limit, we can use L'Hôpital's Rule, which states that if we have a limit of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], we can differentiate the numerator and the denominator until the limit can be evaluated directly.
First, let's verify that the limit is indeed a [tex]\( \frac{0}{0} \)[/tex] form:
[tex]\[ \lim_{x \to 0} (1 - \cos 6x) = 1 - \cos 0 = 0 \][/tex]
[tex]\[ \lim_{x \to 0} 4x^2 = 4 \cdot 0^2 = 0 \][/tex]
Since it is a [tex]\( \frac{0}{0} \)[/tex] form, we can apply L'Hôpital's Rule:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} = \lim_{x \to 0} \frac{d}{dx} \left(1 - \cos 6x\right) \bigg/ \frac{d}{dx} (4x^2) \][/tex]
Now, we differentiate the numerator and the denominator:
[tex]\[ \frac{d}{dx} (1 - \cos 6x) = 6 \sin 6x \][/tex]
[tex]\[ \frac{d}{dx} (4x^2) = 8x \][/tex]
Applying L'Hôpital's Rule, we get:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} \][/tex]
We observe that this is still a [tex]\( \frac{0}{0} \)[/tex] form, so we apply L'Hôpital's Rule again:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} = \lim_{x \to 0} \frac{d}{dx} (6 \sin 6x) \bigg/ \frac{d}{dx} (8x) \][/tex]
Differentiating again:
[tex]\[ \frac{d}{dx} (6 \sin 6x) = 36 \cos 6x \][/tex]
[tex]\[ \frac{d}{dx} (8x) = 8 \][/tex]
Thus, the limit becomes:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} \][/tex]
Evaluating the limit as [tex]\( x \to 0 \)[/tex]:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} = \frac{36 \cdot \cos 0}{8} = \frac{36 \cdot 1}{8} = \frac{36}{8} = \frac{9}{2} \][/tex]
Therefore, for the function [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex], we need:
[tex]\[ A = \frac{9}{2} \][/tex]
So, the value of [tex]\( A \)[/tex] is [tex]\( \frac{9}{2} \)[/tex].
Given the function:
[tex]\[ f(x) = \begin{cases} \frac{1 - \cos 6x}{4x^2} & \text{if } x \neq 0 \\ A & \text{if } x = 0 \end{cases} \][/tex]
To find [tex]\( A \)[/tex], we evaluate the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} \][/tex]
To determine this limit, we can use L'Hôpital's Rule, which states that if we have a limit of the form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], we can differentiate the numerator and the denominator until the limit can be evaluated directly.
First, let's verify that the limit is indeed a [tex]\( \frac{0}{0} \)[/tex] form:
[tex]\[ \lim_{x \to 0} (1 - \cos 6x) = 1 - \cos 0 = 0 \][/tex]
[tex]\[ \lim_{x \to 0} 4x^2 = 4 \cdot 0^2 = 0 \][/tex]
Since it is a [tex]\( \frac{0}{0} \)[/tex] form, we can apply L'Hôpital's Rule:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos 6x}{4x^2} = \lim_{x \to 0} \frac{d}{dx} \left(1 - \cos 6x\right) \bigg/ \frac{d}{dx} (4x^2) \][/tex]
Now, we differentiate the numerator and the denominator:
[tex]\[ \frac{d}{dx} (1 - \cos 6x) = 6 \sin 6x \][/tex]
[tex]\[ \frac{d}{dx} (4x^2) = 8x \][/tex]
Applying L'Hôpital's Rule, we get:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} \][/tex]
We observe that this is still a [tex]\( \frac{0}{0} \)[/tex] form, so we apply L'Hôpital's Rule again:
[tex]\[ \lim_{x \to 0} \frac{6 \sin 6x}{8x} = \lim_{x \to 0} \frac{d}{dx} (6 \sin 6x) \bigg/ \frac{d}{dx} (8x) \][/tex]
Differentiating again:
[tex]\[ \frac{d}{dx} (6 \sin 6x) = 36 \cos 6x \][/tex]
[tex]\[ \frac{d}{dx} (8x) = 8 \][/tex]
Thus, the limit becomes:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} \][/tex]
Evaluating the limit as [tex]\( x \to 0 \)[/tex]:
[tex]\[ \lim_{x \to 0} \frac{36 \cos 6x}{8} = \frac{36 \cdot \cos 0}{8} = \frac{36 \cdot 1}{8} = \frac{36}{8} = \frac{9}{2} \][/tex]
Therefore, for the function [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex], we need:
[tex]\[ A = \frac{9}{2} \][/tex]
So, the value of [tex]\( A \)[/tex] is [tex]\( \frac{9}{2} \)[/tex].
