proof: let e>0 be given. we need to find a d>0, such that lf(x) - 2l<e, when 0<l
ID: 2941195 • Letter: P
Question
proof: let e>0 be given. we need to find a d>0, such that lf(x) - 2l<e, when0<lx - 1/2l<d. a useful algebraic manipulation is lf(x) - 2l = l1/x - 2l = l(1 - 2x)/xl =l2/xl lx -(1/2)l. since we want x to be close to 1/2, we can restrict our deliberlations to lx - 1/2l<1 and see what that says about l2/xl. we would like to find a number M>0. (M=4 worked in the previous ex.), such that l2/xl<M, ie, such that 2<Mlxl, when lx-1/2l<1, unfortunately, this does not work. in fact, lx-1/2l<1 is satisfied by values of x that are arbitraraily close to 0 and 2<Mlxl fails whenlxl<2/M. repair the proof by considering lx-1/2l<1/4 in place of lx - 1/2l. proof: let e>0 be given. we need to find a d>0, such that lf(x) - 2l<e, when
0<lx - 1/2l<d. a useful algebraic manipulation is lf(x) - 2l = l1/x - 2l = l(1 - 2x)/xl =l2/xl lx -(1/2)l. since we want x to be close to 1/2, we can restrict our deliberlations to lx - 1/2l<1 and see what that says about l2/xl. we would like to find a number M>0. (M=4 worked in the previous ex.), such that l2/xl<M, ie, such that 2<Mlxl, when lx-1/2l<1, unfortunately, this does not work. in fact, lx-1/2l<1 is satisfied by values of x that are arbitraraily close to 0 and 2<Mlxl fails whenlxl<2/M. repair the proof by considering lx-1/2l<1/4 in place of lx - 1/2l.
Explanation / Answer
|f(x) - 2| = |2/x||x-(1/2)|. Now we need to find a d>0 such that when |x - (1/2)|<d. To fix the suggested proof, we need to bound |2/x|. We cannot bound 2/x as x gets closer to 0 but note that since we are interested in x close to 0, we can further concentrate on the situation when 1/4 < x < 3/4. In particular, this gives 4/3 < 1/x < 4 => 8/3 < 2/x < 8. Hence under this restriction, |2/x|<8. Hence if we have d = e/8, then if |x - (1/2)| , d we have|f(x) - 2| = |2/x||x - (1/2)| < 8d < e. And that completes the proof.
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