Evaluate 3+jk+k^3 when j=2 and k=6
1 answer:
You might be interested in
The first limit is known as a left-hand limit. We're approaching x = 3 from the left. This means we start with values smaller than x = 3, say x = 2.5 and move closer to x = 3.
Because x is starting off smaller than 3, we use the first piece of the piecewise function. This is the piece that corresponds to x < 3. So we plug x = 3 into that piece to get
2x^2 - x = 2(3)^2 - 3 = 2(9) - 3 = 18 - 3 = 15
So the result for the left-hand limit is 15.
----------------------------------------------------------
The right hand limit will have us start on the other side of x = 3. We start at x = 3.5 and move closer to x = 3. So we'll use the
portion.
Plug x = 3 into the second piece to get
3 - x = 3 - 3 = 0
--------------------------------------------------------------------------
The result of the left-hand limit was 15
The result of the right-hand limit was 0
The fact that the results do not match up means that the overall limit at x = 3 does NOT exist.
Because x is starting off smaller than 3, we use the first piece of the piecewise function. This is the piece that corresponds to x < 3. So we plug x = 3 into that piece to get
2x^2 - x = 2(3)^2 - 3 = 2(9) - 3 = 18 - 3 = 15
So the result for the left-hand limit is 15.
----------------------------------------------------------
The right hand limit will have us start on the other side of x = 3. We start at x = 3.5 and move closer to x = 3. So we'll use the
Plug x = 3 into the second piece to get
3 - x = 3 - 3 = 0
--------------------------------------------------------------------------
The result of the left-hand limit was 15
The result of the right-hand limit was 0
The fact that the results do not match up means that the overall limit at x = 3 does NOT exist.