Question

Given lim x->4 (x/10+1)=7/5, illustrate the definition of the limit by finding delta-epsilon when epsilon = 1/15. Now, I don't expect anybody to go through the whole proof but I don't know what to do in this case because I already know what epsilon is equal to. If somebody could help me with this little step it would greatly appreciated.

Answer 1

start with the definition of the limit being \(\frac{7}{5}\)

Answer 2

that means given \(\epsilon>0\) we can find a \(\delta)\) so that if \(|x-4|<\delta\) then
\[|\frac{x}{10}+1-\frac{7}{4}|<\epsilon\]

Answer 3

in general, \(\delta\) is an expression written in terms of \(\epsilon\)
but in your case \(\epsilon\) is a number, namely \(\frac{1}{15}\) so your \(\delta\) will be a number

Answer 4

Did you mean \[\frac{ x }{ 10 }+\frac{ 7 }{ 5 }\]

Answer 5

no

Answer 6

your expression is \(\frac{x}{10}+1\) right?

Answer 7

and you want to show that the limit is \(\frac{7}{5}\)

Answer 8

that means that the difference between the two can be made arbitrarily small, i.e that
\[|\frac{x}{10}+1-\frac{7}{5}|\] is small

Answer 9

Ok, got it.

Answer 10

now it is some algebra

Answer 11

The hard part =).

Answer 12

you need to solve for \(\delta\) to make
\[|\frac{x}{10}-\frac{2}{5}|<\frac{1}{15}\]

Answer 13

you need to solve for \(\delta\) to make
\[|\frac{x}{10}-\frac{2}{5}|<\frac{1}{15}\]

Answer 14

\[|\frac{x-4}{10}|<\frac{1}{15}\]
\[\frac{1}{10}|x-4|<\frac{1}{15}\]
\[|x-4|<\frac{10}{15}=\frac{2}{3}\]

Answer 15

The part I am struggling with now is the actual proof.

Answer 16

sorry site froze up on me. but you can see that the algebra is not so bad

Answer 17

you mean the proof with \(\epsilon\) instead of \(\frac{1}{15}\) ?

Answer 18

Kind of in reverse. The proof with 1/15 in place of explicitly stating e.

Answer 19

This is probly more trouble for you that it is worth and I apologize for making this difficult.

Answer 20

that one i wrote in detail above

Answer 21

we already have \(\delta=\frac{3}{2}\) works
now run the steps backwards

Answer 22

if
\[|x-4|<\frac{2}{3}\] then
\[\frac{1}{10}|x-4|<\frac{1}{15}\] and so
\[|\frac{x-4}{10}|<\frac{1}{15}\]
\[|\frac{x}{10}-\frac{4}{10}|<\frac{1}{15}\]
\[|\frac{x}{10}+1-\frac{7}{5}|<\frac{1}{15}\]