We need to prove that the following function is continuous at \( x=1 \). The given function is \[ f(x) = \begin{cases} x + 2 & \text{if } x < 1, \\ 3 & \text{if } x = 1, \\ x^2 & \text{if } x > 1. \end{cases} \]
Solution:
For a function \( f(x) \) to be continuous at a point \( x = c \), the following three conditions must be satisfied:
- \( f(c) \) is defined;
- \( \lim_{x \to c} f(x) \) exists;
- \( \lim_{x \to c} f(x) = f(c) \).
1) \( f(1) = 3 \). Therefore, \( f(1) \) is defined.
2) Compute the left-hand and right-hand limits:
\( \lim_{x \to 1^-} f(x) \\ = \lim_{x \to 1^-} (x + 2) = 3\\ \lim_{x \to 1^+} f(x) \\ = \lim_{x \to 1^+} x^2 = 1 \\ \)Since \( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \), the limit \( \lim_{x \to 1} f(x) \) does not exist.
Since the limit \( \lim_{x \to 1} f(x) \) does not exist, \( f(x) \) is not continuous at \( x = 1 \).
Please let me know in the comments if you find any error in this solution.