For which of the following choices?

Question

Let f(x)=3x+1 for all real x and let ε>0. For which of the following choices of δ is abs(f(x)-7)<ε whenever abs(x-2)<δ?

 

a) ε/4

b) ε/2

c) ε/(ε+1)

d) (ε+1)/ε

e) 3ε

 

Answer: A

Answer 1

Since the slope of  abs(f(x) -7) is ± 3,    The straightforward boundary  value of δ is δ = ε/3.

 

Thus a) is ok since it is less than this boundary value.

         b)   is not ok since it is greater

         c)  is ok if  ε >2, but not in general   , so not ok overall

           d) is not ok for similar reasons as c)

           e)  is not ok   since 3ε > ε/3

 

Thus the only good answer is a)

Answer 2

abs(f(x)-7)<ε means:

abs((3x+1)-7) = abs(3x-6) = 3 abs(x-2) < ε, or

abs(x-2) < ε/3

Any δ ≤ δ_max= ε/3 will give abs(f(x)-7)<ε whenever abs(x-2)<δ.

Of the five choices given, only  ε/4 ≤ ε/3 for all ε>0, so the answer is A.

 

Note: this is equivalent to using the definition of limit to show that lim_x→2 (3x+1) = 7.