Using the delta-epsilon method, prove that the limit as x approaches 2 of 1/x is equal to 1/2.

Question

When finding delta, use the absolute value of x-2 is less than 2 so that delta=min{2,__} 

 

 

This is what I've done so far:

 

f(x)=1/x

L=1/2

a=2

 

Find delta:

absolute value of (1/x-1/2)= absolute value of [(2/2)(1/x)-(x/x)(1/2)]= absolute value of [(2-x)/2x]= 1/(absolute value of 2x)*[absolute value of (2-x)]

 

Assume 1/2x is bounded by some constant "c"

1/(absolute value of 2x)=c thus, 1/(absolute value of 2x)*[absolute value of (x-2)]<c*[absolute value of (x-2)]<epsilon

 

Restrict x so that it is within "2" (it has to be 2 and this is where it gets tricky) away from a=2

[absolute value of (x-2)]<2

this implies that -2<x-2<2 = 0<x<4 = 0<2x<8 =0<1/8<1/2x

 

This is where I got stuck and I don't know where to go from here!

Answer 1

I think the pickle you're in is that you're trying to find a number c such that 1/|2x| < c but when you assumed that |x-2| < 2 you got that 1/8 < 1/|2x| < infty.  That is, 1/|2x| is unbounded for 0 < x < 4 and for that matter so is |x-2|/|2x| = |1/x - 1/2|.  So if δ = 2, then |1/x - 1/2| isn't bounded by any ε > 0.

 

Perhaps the reason for the restriction to stay within 2 units of 2 is because the function 1/x is not defined at x = 0.  

 

Instead of 2 assume that |x-2| < 1.  Then 1/|2x| < 1/2.  This implies that |x-2|/|2x| < 1/2*|x-2|.  So take δ = min{1,2ε}.