Whether you're a student or a professional, IDNLearn.com has answers for everyone. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.
Sagot :
To determine the value of [tex]\(\lambda\)[/tex] that makes the function [tex]\(f(x)\)[/tex] continuous at [tex]\(x = 0\)[/tex], we need to ensure that the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 0 is equal to the value of the function at [tex]\(x = 0\)[/tex]. Here, the function is defined as:
[tex]\[ f(x) = \begin{cases} \frac{1 - \cos(x)}{x^2} & \text{if } x \neq 0 \\ \lambda & \text{if } x = 0 \end{cases} \][/tex]
The function [tex]\(f(x)\)[/tex] is continuous at [tex]\(x = 0\)[/tex] if:
[tex]\[ \lim_{x \to 0} f(x) = f(0) \][/tex]
Since [tex]\(f(0) = \lambda\)[/tex], we need to find the limit of [tex]\(\frac{1 - \cos(x)}{x^2}\)[/tex] as [tex]\(x\)[/tex] approaches 0.
Let's evaluate this limit step by step:
1. Expression Analysis:
[tex]\(\frac{1 - \cos(x)}{x^2}\)[/tex] is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex] as [tex]\(x \to 0\)[/tex], so we can use L'Hôpital's rule to find the limit.
2. First Application of L'Hôpital's Rule:
L'Hôpital's rule states that if the limit of [tex]\(\frac{f(x)}{g(x)}\)[/tex] is of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists.
In our example:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \][/tex]
The numerator is [tex]\(1 - \cos(x)\)[/tex] and the denominator is [tex]\(x^2\)[/tex]. We differentiate the numerator and denominator with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}[1 - \cos(x)] = \sin(x) \][/tex]
[tex]\[ \frac{d}{dx}[x^2] = 2x \][/tex]
Applying L'Hôpital's rule:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{\sin(x)}{2x} \][/tex]
3. Second Application of L'Hôpital's Rule:
The limit [tex]\(\lim_{x \to 0} \frac{\sin(x)}{2x}\)[/tex] is still of the form [tex]\(\frac{0}{0}\)[/tex], so we apply L'Hôpital's rule again.
Differentiate the numerator and denominator once more:
[tex]\[ \frac{d}{dx}[\sin(x)] = \cos(x) \][/tex]
[tex]\[ \frac{d}{dx}[2x] = 2 \][/tex]
Applying L'Hôpital's rule again:
[tex]\[ \lim_{x \to 0} \frac{\sin(x)}{2x} = \lim_{x \to 0} \frac{\cos(x)}{2} = \frac{\cos(0)}{2} = \frac{1}{2} \][/tex]
4. Conclusion:
Thus,
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \][/tex]
To make [tex]\(f(x)\)[/tex] continuous at [tex]\(x = 0\)[/tex], we set [tex]\(\lambda\)[/tex] to be equal to this limit. Therefore, the value of [tex]\(\lambda\)[/tex] is:
[tex]\[ \lambda = \frac{1}{2} \][/tex]
[tex]\[ f(x) = \begin{cases} \frac{1 - \cos(x)}{x^2} & \text{if } x \neq 0 \\ \lambda & \text{if } x = 0 \end{cases} \][/tex]
The function [tex]\(f(x)\)[/tex] is continuous at [tex]\(x = 0\)[/tex] if:
[tex]\[ \lim_{x \to 0} f(x) = f(0) \][/tex]
Since [tex]\(f(0) = \lambda\)[/tex], we need to find the limit of [tex]\(\frac{1 - \cos(x)}{x^2}\)[/tex] as [tex]\(x\)[/tex] approaches 0.
Let's evaluate this limit step by step:
1. Expression Analysis:
[tex]\(\frac{1 - \cos(x)}{x^2}\)[/tex] is an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex] as [tex]\(x \to 0\)[/tex], so we can use L'Hôpital's rule to find the limit.
2. First Application of L'Hôpital's Rule:
L'Hôpital's rule states that if the limit of [tex]\(\frac{f(x)}{g(x)}\)[/tex] is of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided the limit on the right exists.
In our example:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \][/tex]
The numerator is [tex]\(1 - \cos(x)\)[/tex] and the denominator is [tex]\(x^2\)[/tex]. We differentiate the numerator and denominator with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}[1 - \cos(x)] = \sin(x) \][/tex]
[tex]\[ \frac{d}{dx}[x^2] = 2x \][/tex]
Applying L'Hôpital's rule:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{\sin(x)}{2x} \][/tex]
3. Second Application of L'Hôpital's Rule:
The limit [tex]\(\lim_{x \to 0} \frac{\sin(x)}{2x}\)[/tex] is still of the form [tex]\(\frac{0}{0}\)[/tex], so we apply L'Hôpital's rule again.
Differentiate the numerator and denominator once more:
[tex]\[ \frac{d}{dx}[\sin(x)] = \cos(x) \][/tex]
[tex]\[ \frac{d}{dx}[2x] = 2 \][/tex]
Applying L'Hôpital's rule again:
[tex]\[ \lim_{x \to 0} \frac{\sin(x)}{2x} = \lim_{x \to 0} \frac{\cos(x)}{2} = \frac{\cos(0)}{2} = \frac{1}{2} \][/tex]
4. Conclusion:
Thus,
[tex]\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \][/tex]
To make [tex]\(f(x)\)[/tex] continuous at [tex]\(x = 0\)[/tex], we set [tex]\(\lambda\)[/tex] to be equal to this limit. Therefore, the value of [tex]\(\lambda\)[/tex] is:
[tex]\[ \lambda = \frac{1}{2} \][/tex]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Thank you for visiting IDNLearn.com. We’re here to provide dependable answers, so visit us again soon.
