Answer :
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]
