Answers:
![\displaystyle \lim_(x \to 2^(+)) f(x) = 1\\\\\displaystyle \lim_(x \to 2^(-)) f(x) = 1\\](https://img.qammunity.org/2021/formulas/mathematics/college/xkspqo1ubn0y6njo5ggwzeax9j1ba834ip.png)
Both result in the same limit value. This allows us to say
without the plus or minus over the 2.
The left and right hand limits may not always match like this.
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Step-by-step explanation:
The notation
means that we are approaching x = 2 from the right hand side. This is from the positive direction. So we start at say x = 3 and move to x = 2.5 then to x = 2.1 then to x = 2.01 and so on.
Because we started with values x > 2, we will use the third definition of the piecewise function
if x > 2, then f(x) = 3x-5
Plug in x = 2 to get
f(x) = 3x-5
f(2) = 3(2)-5
f(2) = 6-5
f(2) = 1
This shows
![\displaystyle \lim_(x \to 2^(+)) f(x) = 1](https://img.qammunity.org/2021/formulas/mathematics/college/jctyqosnq98qlk58lwo52ri3tkq1odmo04.png)
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For the other limit, we're approaching x = 2 from the negative side. So we could start at say x = 0, then move to x = 1, then to x = 1.5 then to x = 1.9 then to x = 1.99, and so on.
We're using x values such that x < 2 now.
So we'll be using the first definition of the piecewise function
If x < 2, then f(x) = x^2 - 3
f(x) = x^2-3
f(2) = 2^2-3
f(2) = 4-3
f(2) = 1
We end up with
![\displaystyle \lim_(x \to 2^(-)) f(x) = 1](https://img.qammunity.org/2021/formulas/mathematics/college/u8i4y91ex15fcjay23r6msall28imih2ur.png)
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Both right hand limit and left hand limit result in the same value
Because
![\displaystyle \lim_(x \to 2^(+)) f(x) = \displaystyle \lim_(x \to 2^(-)) f(x) = 1](https://img.qammunity.org/2021/formulas/mathematics/college/jxh8apgulbdryiq49se3oqt6kdxgbyi1tj.png)
We can shorten that to
meaning we can approach x = 2 from either direction to arrive at the same limiting value.
A thing to notice is that f(2) is not equal to 1. Instead the second line of the piecewise function says f(2) = 3.
The fact that the limit as x approaches 2 and f(2) don't agree means this function is not continuous at x = 2.
The graph shows this. We have a removable discontinuity where we effectively picked the point off the graph and move it upward.
See the diagram below.