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Lex M.

asked • 04/09/15

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

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!

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