To analyze the continuity of the function [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex], we need to determine if the left-hand limit as [tex]\( x \)[/tex] approaches 0, the right-hand limit as [tex]\( x \)[/tex] approaches 0, and the value of the function at [tex]\( x = 0 \)[/tex] are all equal.
Given the function:
[tex]\[
f(x) = \begin{cases}
\frac{e^x - 1 - x}{x^2}, & \text{if } x \neq 0 \\
\frac{1}{2}, & \text{if } x = 0
\end{cases}
\][/tex]
1. Compute the left-hand limit as [tex]\( x \to 0 \)[/tex]:
We need to find the limit of [tex]\( \frac{e^x - 1 - x}{x^2} \)[/tex] as [tex]\( x \)[/tex] approaches 0. Therefore, we calculate:
[tex]\[
\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}
\][/tex]
This limit can be evaluated using L'Hôpital's rule. Applying L'Hôpital's rule twice (since both the numerator and denominator approach 0 as [tex]\( x \to 0 \)[/tex]):
First application:
[tex]\[
\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}
\][/tex]
Thus, we have:
[tex]\[
\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \frac{1}{2}
\][/tex]
2. Determine the right-hand limit at [tex]\( x = 0 \)[/tex]:
From the definition of the function [tex]\( f \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = \frac{1}{2}
\][/tex]
3. Check the continuity condition:
To check the continuity at [tex]\( x = 0 \)[/tex], we need to verify if:
[tex]\[
\lim_{x \to 0} f(x) = f(0)
\][/tex]
From our computations:
[tex]\[
\lim_{x \to 0} f(x) = \frac{1}{2} = f(0)
\][/tex]
Since both the left-hand limit and the right-hand limit as [tex]\( x \)[/tex] approaches 0 are equal to [tex]\( f(0) \)[/tex], we conclude that [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex].
Thus, the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex].
