annual fee is your awnser hope it helped
its b. hope this helps i guess
First .... the limit of the given function is NOT EQUAL to 5 (it diverges so it has no limit). There is a typo. The 14 should be positive.
![\lim_{x \to\ 7} \bigg(\dfrac{x^2-9x+14}{x-7}\bigg)=5](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%5C%207%7D%20%5Cbigg%28%5Cdfrac%7Bx%5E2-9x%2B14%7D%7Bx-7%7D%5Cbigg%29%3D5)
The precise definition of a limit is:
![\text{If for every } \epsilon>0\text{ there exists a }\delta >0\text{ such that}\\|f(x)-L|](https://tex.z-dn.net/?f=%5Ctext%7BIf%20for%20every%20%7D%20%5Cepsilon%3E0%5Ctext%7B%20there%20exists%20a%20%7D%5Cdelta%20%3E0%5Ctext%7B%20such%20that%7D%5C%5C%7Cf%28x%29-L%7C%3C%5Cepsilon%5C%20%5Ctext%7Bwhenever%20%7D%7Cx-a%7C%3C%5Cdelta)
Given:
![f(x) = \dfrac{x^2-9x+14}{x-7}\\\\L=5\\\\a=7\\\\\\|f(x)-L|](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cdfrac%7Bx%5E2-9x%2B14%7D%7Bx-7%7D%5C%5C%5C%5CL%3D5%5C%5C%5C%5Ca%3D7%5C%5C%5C%5C%5C%5C%7Cf%28x%29-L%7C%3C%5Cepsilon%5Cqquad%20%5Cqquad%20%5Ctext%7Bwhenever%7D%5Cquad%20%7Cx-a%7C%3C%5Cdelta%5C%5C%5C%5C%5Cbigg%7C%5Cdfrac%7Bx%5E2-9x%2B14%7D%7Bx-7%7D-5%5Cbigg%7C%3C%5Cepsilon%5Cqquad%20%5Ctext%7Bwhenever%7D%5Cquad%20%7Cx-7%7C%3C%5Cdelta%5C%5C%5C%5C%5C%5C%5Cbigg%7C%5Cdfrac%7B%28x-2%29%28x-7%29%7D%7Bx-7%7D-5%5Cbigg%7C%3C%5Cepsilon%5Cqquad%20%5Ctext%7Bwhenever%7D%5Cquad%20%7Cx-7%7C%3C%5Cdelta%5C%5C%5C%5C%5C%5C%7Cx-2-5%7C%3C%5Cepsilon%5Cqquad%20%5Cqquad%20%5Cquad%20%5Ctext%7Bwhenever%7D%5Cquad%20%7Cx-7%7C%3C%5Cdelta%5C%5C%5C%5C%7Cx-7%7C%3C%5Cepsilon%5Cqquad%20%5Cqquad%20%5Ctext%7Bwhenever%7D%5Cquad%20%7Cx-7%7C%3C%5Cdelta)
⇒ ![\epsilon = \delta](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%5Cdelta)
When ε = 0.1, δ = 0.1
When ε = 0.01, δ = 0.01
15/3 + 4/3 = 19/3
All you have to do is subtract 4/3 from 19/3 to get 15/3