To determine the value of [tex]\(\lambda\)[/tex] such that the function
[tex]\[ f(x) = \begin{cases}
\frac{1 - \cos x}{x^2}, & \text{if } x \neq 0 \\
\lambda, & \text{if } x = 0
\end{cases} \][/tex]
is continuous at [tex]\( x = 0 \)[/tex], we need to ensure that the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex] is equal to [tex]\( f(0) \)[/tex]. Mathematically, this condition for continuity at [tex]\( x = 0 \)[/tex] is expressed as:
[tex]\[ \lim_{x \to 0} f(x) = f(0) = \lambda \][/tex]
Hence, we need to calculate the limit:
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \][/tex]
To evaluate this limit, let's recognize that direct substitution of [tex]\( x = 0 \)[/tex] leads to an indeterminate form [tex]\( \frac{0}{0} \)[/tex]. Hence, we use L'Hôpital's Rule, which is suitable for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex]. L'Hôpital's Rule states that:
[tex]\[ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \][/tex]
First, let's identify [tex]\( f(x) = 1 - \cos x \)[/tex] and [tex]\( g(x) = x^2 \)[/tex].
We then find their derivatives:
[tex]\[ f'(x) = \frac{d}{dx}(1 - \cos x) = \sin x \][/tex]
[tex]\[ g'(x) = \frac{d}{dx}(x^2) = 2x \][/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]
Now, we can use the known limit [tex]\(\lim_{x \to 0} \frac{\sin x}{x} = 1\)[/tex]:
[tex]\[ \lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2} \][/tex]
Thus,
[tex]\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \][/tex]
So, to ensure [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex], we set the value of [tex]\(\lambda\)[/tex] equal to this limit:
[tex]\[ \lambda = \frac{1}{2} \][/tex]
Therefore, the value of [tex]\(\lambda\)[/tex] for the function [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 0 \)[/tex] is:
[tex]\[ \boxed{\frac{1}{2}} \][/tex]
