For any >0, suppose that there is a >0 so that the implication below is true:...

 

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For any >0, suppose that there is a >0 so that the implication below is true: a. If x=0.1, find the largest value for >0 which ensures the above implication is true. 8= b. If =0.01, find the largest value for >0 which ensures the above implication is true. 8= if 0<x-71<6, then 12√x+9-81 <8 c. Using the work you did in parts a. and b., find the largest value for >0 which ensures the above implication is true for an arbitrary >0. (Your answer should be an expression involving > .) 6= d. The result you found in part c. proves that a certain limit exists. Fill in the blanks below for the pieces of the corresponding limit. i. The value x is approaching is a= ii. The function is f(x)= iii. The value of the limit is L= For any >0, suppose that there is a >0 so that the implication below is true: a. If x=0.1, find the largest value for >0 which ensures the above implication is true. 8= b. If =0.01, find the largest value for >0 which ensures the above implication is true. 8= if 0<x-71<6, then 12√x+9-81 <8 c. Using the work you did in parts a. and b., find the largest value for >0 which ensures the above implication is true for an arbitrary >0. (Your answer should be an expression involving > .) 6= d. The result you found in part c. proves that a certain limit exists. Fill in the blanks below for the pieces of the corresponding limit. i. The value x is approaching is a= ii. The function is f(x)= iii. The value of the limit is L=

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a If 01 then the largest that works is 01 2x98 2798 2168 212 01 Taking 01 satisfies the above ineq... View the full answer