Question

Prove that
lim as x approaches 2 (1/itionx) = 1/2...is there anotehr way besides direct substitution

Answer 1

lim x approaches 2 (1/x) = 1/x

Answer 2

i want to prove it w. epsiolon and delta

Answer 3

we can probably do this if you like

Answer 4

no problem

Answer 5

first i assume you mean
\[\lim_{x\rightarrow 2}\frac{1}{x}=\frac{1}{2}\]

Answer 6

so you have to show that given any epsilon greater than zero there will be some delta (which you write in terms of epsilon) so that if
\[|x-2|<\delta\] then
\[|\frac{1}{x}-\frac{1}{2}|<\epsilon\]

Answer 7

as usual we work backwards to see what we need.

Answer 8

\[|\frac{1}{x}-\frac{1}{2}|<\epsilon\]
\[|\frac{2-x}{2x}|<\epsilon\]
\[\frac{|2-x|}{|2x|}<\epsilon\] and now we see that we are almost there because
\[|x-2|=|2-x|\] and that is the one we have control over

Answer 9

the only thing we don't have a bound for is
\[|2x|\] so we do the usual trick and make a bound for
\[|x-2|\] say make sure that
\[|x-2|<1\] which we are allowed since we control that term

Answer 10

if
\[|x-2|<1\] then
\[-1<x-2<1\] so
\[1<x<3\]
and now we know that the smallest the denominator can be is 1

Answer 11

or rather 2, since it is
\[|2x|\] so we are in good shape now

Answer 12

b4 we move on how is |x−2|=|2−x|

Answer 13

if the smallest the denominator can be is 2, then
\[\frac{|2-x|}{|2x|}\leq2|x-2|\] so make sure that and so if we make
\[|x-2|<\frac{\epsilon}{2}\] then
\[|\frac{2-x}{2x}|\leq 2|x-2|<2\frac{\epsilon}{2}=\epsilon\]and we are done

Answer 14

it is always the case that
\[|b-a|=|a-b|\]

Answer 15

you can try it with numbers to see, but this says the distance between a and b is the same as the distance between b and a

Answer 16

oh ok..i see cuz its absoute value

Answer 17

right

Answer 18

so if it equals epsilon at te hend taht means teh limit is proved or taht it actually esists?

Answer 19

i think i made a mistake actually but you can always err on the side of caution. in fact if |2x|>2 then
\[\frac{|2-x|}{|2x|}<\frac{1}{2}|x-2|\] so we could have picked a different delta, but as long as we get something smaller that is ok

Answer 20

yes so if you can arrange it so that |x-a|< something insures that |f(x)-L|< epsilon, then you are done