Question

Calculus piecewise function. ​

Answer 1

Part A

The notation
`\lim_(x \to 2^(+))f(x)` means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows
`\lim_(x \to 2^(+))f(x) = 2`

Then for the left hand limit
`\lim_(x \to 2^(-))f(x)`, we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore,
`\lim_(x \to 2^(-))f(x) = 1`

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Part B

Because
`\lim_(x \to 2^(+))f(x) \\e \lim_(x \to 2^(-))f(x)` this means that the limit
`\lim_(x \to 2)f(x)` does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

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Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

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Part D

As mentioned earlier, since
`\lim_(x \to 4^(+))f(x) = \lim_(x \to 4^(-))f(x) = 3`, this means the limit
`\lim_(x \to 4)f(x)` does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.