Consider the following limit. lim x→2 (3x + 2) Find the limit L.

Question

Well I know that the limit L is equal to eight. I know I get 8 by plugging in the 2 into 3x+2, but then it asks me the following:

 

(a) Find δ > 0 such that |f(x) − L| < 0.01 whenever 0 < |x − c| < δ. (Round your answer to five decimal places.)

 

(b) Find δ > 0 such that |f(x) − L| < 0.005 whenever 0 < |x − c| < δ. (Round your answer to five decimal places.)

 

What is this trying to tell me to do? I am a very lost and would like to better understand the wording, or rather the lack of wording I can't decipher. Thank you! 

 

Answer 1

a.  lim_x→c f(x) = L    So, c = 2, f(x) = 3x+2, and L = 8.

 

     Find δ>0 such that l (3x+2) - 8 l < 0.01 whenever 0 < l x-2 l < δ:

 

         Let δ = 0.01 / 3 = 0.00333    

 

         If 0 < l x-2 l < 0.01/3, then -0.01/3 < x-2 < 0.01/3

 

                                                      -0.01 < 3x - 6 < 0.01

 

                                                      -0.01 < (3x+2) - 8 < 0.01

 

                                                      -0.01 < f(x) - L < 0.01

 

                                                      l f(x) - L l < 0.01

 

For part b, let δ = 0.005/3 and proceed as in part a.