Explanation:
For a. We must find the lim as x approaches 2 from the right.
We can use direct substitution for the second equation because
- We can direct subsitute since we not dealing with no asymptotes
- We use the second equation because that is when x is greater than or equal to two so we use that because we can determine the limit.
So just subsitue 2 in for x.
![- 6(2) {}^(2) + 2m](https://img.qammunity.org/2023/formulas/mathematics/college/wxjj2pzgpxnwg4vca419rio26xqhl7etii.png)
![- 6(4) + 2m](https://img.qammunity.org/2023/formulas/mathematics/college/k0vmh6cf8xq5ngz0n43q2kwg8p3ts0s1en.png)
![2m - 24](https://img.qammunity.org/2023/formulas/mathematics/college/g5m62nn6of80fetbhxbw9zgn3tqjhhzuas.png)
So the limit as x approaches 2 from the right, is 2m-24.
b. We now must find the limit as x approaches 2 from the left,
This time we will use the top equation because since the x values for the top equation is only defined to be less than 2, we can know the behavior as it approaches 2 from the left.
Here we direct subsitue again
![6(2) + m](https://img.qammunity.org/2023/formulas/mathematics/college/6ptxoc5xr4b5aaq5hgt7q7j0guojhyumvt.png)
![12 + m](https://img.qammunity.org/2023/formulas/mathematics/college/isntmbykkn2dnqjb7v3lz2kqqi96qze8sv.png)
So as x approaches 2 from the left, the limit is 12+m
Here we let
![12 + m = 2m - 24](https://img.qammunity.org/2023/formulas/mathematics/college/nf70wkiubz4kcem25rh36wdbskmldf4xii.png)
![36 = m](https://img.qammunity.org/2023/formulas/mathematics/college/fb69r7ij2srcarb1p1pgnjxeamdqqxpymv.png)
The value is 36 so
m=36