If . For what value (s) of does exists?
For a < 0, the function reduces to f(x) = |x| + 1 = -x + 1, which is a linear function, hence continuous. Therefore, lim_{x→a} f(x) exists for all a < 0. Similarly, for a > 0, f(x) = |x| - 1 = x - 1, which is also linear and continuous, so the limit exists for all a > 0. At a = 0, we compute the left-hand limit as x→0⁻: f(x) = |x| + 1 → 1, and the right-hand limit as x→0⁺: f(x) = |x| - 1 → -1. Since these limits are not equal, lim_{x→0} f(x) does not exist.
Hence, the limit exists for all real numbers a except a = 0.
Explanation
The solution follows the concept from the textbook that the limit exists at a point if the left-hand and right-hand limits coincide. For a<0 and a>0, the function is linear in a neighbourhood of a, so the limit exists. At a=0, the left-hand limit (1) and right-hand limit (-1) differ, similar to Illustration 9 and Question 26 in the context, where the limit does not exist due to unequal one-sided limits.
Solution Steps
Step 1: Consider three cases: a < 0, a > 0, and a = 0.
Step 2: For a < 0, both sides of a are negative, so f(x) = |x| + 1 = -x + 1, a linear function, hence continuous. Thus limit exists.
Step 3: For a > 0, both sides of a are positive, so f(x) = |x| - 1 = x - 1, linear and continuous. Thus limit exists.
Step 4: For a = 0, compute LHL: lim_{x→0⁻} f(x) = lim_{x→0⁻} (|x|+1) = 1. Compute RHL: lim_{x→0⁺} f(x) = lim_{x→0⁺} (|x|-1) = -1. Since LHL RHL, limit does not exist.
Step 5: Conclude that limit exists for all a 0.