Consider the piecewise functionf (x) =x + 1, if x < −21, if − 2 ≤ x ≤ 1x2, if x > 1.(i) Find limx→−2 f (x) if it exists
Question
Consider the piecewise function
f (x) =
\begin{cases}
x + 1, & \text{if } x < -2 \
1, & \text{if } -2 \leq x \leq 1 \
x^2, & \text{if } x > 1.
\end{cases}
(i) Find if it exists
Solution
The limit of a function as x approaches a certain value is the value that the function approaches as x gets closer and closer to that value.
In this case, we are asked to find the limit of the function f(x) as x approaches -2.
The function f(x) is defined as a piecewise function, which means it has different definitions for different ranges of x.
For x < -2, f(x) = x + 1. For -2 ≤ x ≤ 1, f(x) = 1. For x > 1, f(x) = x^2.
Since we are looking for the limit as x approaches -2, we need to consider the definition of the function for x values close to -2.
Looking at the definitions of the function, we see that as x approaches -2, the function is defined as 1 (since -2 ≤ x ≤ 1).
Therefore, the limit of the function as x approaches -2 is 1.
So, lim(x→-2) f(x) = 1.
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